a greater lift leads to a gain in altitude
Here's where you've first gone astray. Lift is actually less than weight in a sustained linear climb. The fundamental thing that makes a sustained steady-state climb possible is that the Thrust vector is pointing upward rather than horizontally, which is only true when Thrust is greater than Drag. We'll return to this point later in this answer.
But why do airplanes climb by "pointing the nose up"?
Regardless of whether we choose to 1) climb at a high (but constant) angle-of-attack and a lower airspeed, or to 2) accelerate to a higher airspeed and climb at a low (but constant) attack, the aircraft will be somewhat nose-high in the climb because the flight path is aimed upwards, and the pitch attitude of the fuselage is the sum of the climb angle of the flight path plus the angle-of-attack of the wing minus the angle-of-incidence (i.e. the "rigging angle" of the wing relative to fuselage).
A third way to climb would be to keep the same pitch attitude that the aircraft had in level (constant-altitude) flight, but this would constrain the angle-of-attack to stay very low-- the higher the climb rate and the steeper the climb path, the lower the angle-of-attack would be forced to go. This is not the kind of feedback loop that leads to high rate of climb!
To understand why, in the artificial situation where the pitch attitude of the aircraft is constrained to be fixed, the angle of the climb path affects the angle-of-attack of the wing, you have to understand that the airflow or "relative wind" felt by an aircraft in flight is exactly opposite in direction to the aircraft's path of travel through the airmass-- which in this case is the path of the climb. (For simplicity we're assuming no wind or updraft/downdraft -- those things can change the climb angle achieved relative to the ground without changing the "relative wind" felt by the airplane, but that's not really what this question was about.) Understanding that the relative wind "felt" by an airplane is always exactly opposite to the airplane's flight path through the airmass, is one of the most important things in understanding how an airplane flies.
Therefore even in an aircraft with an unusually high angle -of-incidence like the B-52, the aircraft will be nose-high in a steep climb.
In theory an aircraft even an aircraft with zero angle-of-incidence could generate lift with the fuselage exactly horizontal. If the flight path were climbing slightly, then the wing would be flying at a slightly negative angle-of-attack, but a cambered airfoil can still create lift in such a situation. But the aircraft would generate a much higher ratio of Lift to Drag if the wing were at some higher angle-of-attack. Even though Lift is less than Weight in a climb, a high ratio of Lift to Drag is still correlated to a steep climb angle. See this related ASE answer to learn why: Does lift equal weight in a climb?
The highest L/D ratios are generated at relatively high angles-of-attack. So this is when we will see the steepest climb angle. The highest climb rate comes at a somewhat lower angle-of-attack, but the aircraft's nose will still be pitched well above the horizon, because of the simple fact that the pitch attitude of the fuselage is the sum of the climb angle of the flight path plus the angle-of-attack of the wing minus the angle-of-incidence of the wing relative to the fuselage.
Is it right that basically an airplane just needs to accelerate to climb?
No, for a steady-state linear climb at a constant airspeed, the aircraft also has to create more Thrust than Drag, and it also has to point the Thrust vector upwards.
At this point we need to revisit the paragraph beginning "A third way to climb would be to keep the same pitch attitude that the aircraft had in level (constant-altitude) flight". There is actually another problem here besides the fact that we'd be forcing the wing to fly at a very low angle-of-attack, where the L/D ratio is poor. The other problem is that the Thrust vector is staying horizontal, and thus a sustained steady-state climb is possible.
(Naturally, we can zoom-climb or even loop a glider with no thrust at all. In loop or zoom climb the requirement for a close vector polygon of Lift, Weight, Drag, and Thrust (if present) vanishes, so the constraints are completely different than in a sustained steady-state climb.)
Consider the case of an aircraft like the B-52. The wing is mounted at a high angle-of-incidence to the fuselage to accommodate the "bicycle" landing gear design by allowing a no-rotation takeoff, and to reduce drag in long-range cruising flight. Even with the fuselage level relative to the airflow, the wing is at an efficient angle-of-attack, with a high L/D ratio. If the aircraft is creating more Lift than its Weight, does this mean that it is established in a steady-state climb? No, it means the flight path will curve or bend upwards, causing the aircraft to pitch upwards, which gives the Thrust vector an upward component. At this point Lift will actually decrease slightly to a value that is smaller than Weight as the aircraft settles into a steady-state climb with Thrust greater than Drag, the nose pointing above horizon, and the Thrust vector pointing upwards and helping to support part of the aircraft's weight.
Note that as we change the angle-of-attack of the wing and change the ratio of Lift coefficient to Drag coefficient, for shallow to moderate climb or descent angles, the airspeed eventually responds in such a way that Lift actually stays almost constant, while Drag varies greatly. The reason we choose an optimum angle-of-attack for climbing is really not to maximize Lift, but rather to minimize Drag and thus maximize the ratio of Thrust to Drag. But regardless of whether we've chosen an angle-of-attack that yields a high L/D ratio or a low L/D ratio, if the Thrust vector is pointing horizontally rather than upward, then we aren't climbing - at least not for more than a brief instant. (More on this later!)
Again, for more on the relationship between Thrust, Drag, Lift, and Weight in a climb, see the related ASE answer Does lift equal weight in a climb?
A closing note-- an exotic situation which is not characteristic of normal free flight (meaning that the aircraft is not connected by a towline to another vehicle that is providing the thrusting force) was discussed in this related ASE question and answer. The situation involves a wing sliding up and down on a pole attached to a cart. In this case, even though the thrust vector can be construed to be horizontal, the wing can indeed climb slowly up the pole while maintaining a constant level pitch attitude, but its angle-of-attack relative to the airflow will be decreased as its climb rate increases, causing a self-limiting effect on the climb rate, as discussed in the present answer.
And now a closing note to the closing note-- earlier, we stated "if the Thrust vector is pointing horizontally rather than upward, then we aren't climbing." We've also noted that a glider can be looped with no Thrust at all. A powered plane can also be "zoom climbed" even if Thrust is less than drag, but the airspeed will be decreasing. Note that during the "zoom climb", the thrust line is still usually pointing upward.
Can we come up with a really contrived case where we "zoom climb" without pitching up at all? Yes we can-- but the climb will be very brief. For example, let's say that we're pulling out of a loop. Let's say that we're "pulling" 4G's-- the Lift vector is four times the aircraft's Weight. Just before we reach a horizontal pitch attitude, the airspeed will typically be decreasing, meaning that Drag is greater than Thrust. As we continue to pull up, there will be an instant in time where the pitch attitude is exactly horizontal, but Lift is still much greater than Weight. At that instant, if we relax the back pressure and move the stick forward as needed to exactly freeze the aircraft's pitch attitude, the flight path will still continue to curve upward for a very brief interval of time, until the upward curve of the flight path decreases the angle-of-attack of the wing to the point where Lift vector is equal to the Weight vector, or more precisely, the point where the Lift vector is equal to the component of the Weight vector that acts perpendicular to the flight path. At that instant the centripetal acceleration is zero. The linear acceleration cannot be zero-- as we continue to hold the pitch attitude of the fuselage constant, the airspeed will decrease, and then the flight path will curve downward again until it is exactly horizontal. When the Thrust vector is exactly horizontal, steady-state flight is only possible in the horizontal direction, not in the upward or downward direction. From the pilot's point of view, what has happened is that we've reach a level pitch attitude and then we've rather rapidly "unloaded" the wing to near 1-G condition and transitioned to approximately horizontal flight. The fact that the aircraft did climb very briefly with the fuselage exactly level would probably be impossible to detect without special instrumentation. But yes, technically, it is possible to achieve a very brief interval of climbing flight with the Thrust vector remaining exactly horizontal, and in fact something close to this happens almost every time we transition from a dive to an horizontal pitch attitude, unless we somehow manage to control the throttle in such a way that the airspeed remains exactly constant during the final portion of the pull-out.
It should be clear by now to the reader that this very brief interval of climbing flight with a fixed horizontal pitch attitude, is not the dynamic that we see during any steady-state climb.