A similar question is:
The answers point to scramjets. Cited reports focus on combustion kinetics.
a block of gas entering the combustion chamber must mix with fuel and have sufficient time for initiation and reaction, all the while traveling supersonically through the combustion chamber, before the burned gas is expanded through the thrust nozzle. This places stringent requirements on the pressure and temperature of the flow, and requires that the fuel injection and mixing be extremely efficient.
For a nuke, no combustion is needed. How does this change the altitude limit?
The document cited by mins 3 concludes that speed and altitude are limited by reactor wall temperature for a nuclear-powered, direct air, shield-less, ram-jet. Twall of 1800 Rankine enables flight at "70,000 feet and a flight Mach number of 3.0, a uranium investment of 81 pounds and a missile gross weight of 64,000 pounds." It goes on to say "lift-drag ratios as low as 3.5 could be tolerated. ... lift-to-drag ratio was estimated at about 5.0."
The nuclear-powered ramjet combines the conventional diffuser and exhaust nozzle with a nuclear reactor. The diffuser decelerates the free-stream air before it enters the reactor passages. Reactor inlet air Mach number is about 0.28.
The question then becomes, for a nuclear scramjet, where inlet air is not braked to subsonic speeds, what improvement in ceiling is attainable?
The conventional fuel-burning scramjet would be limited by the oxygen available at altitude. The nuke needs only atmospheric mass. There is plenty of air at 81,000 feet to provide the needed thrust, according to my amateur calcuations:
W = 64,000 lbf, L/D = 4, let L = W so W/D = 4
drag D = W/4 = 64,000/4 = 16,000 lbf
We need thrust T = D
T = mdot*(u - v) = (mass flow rate)(exhaust velocity - intake velocity)
For a nuclear thermal rocket Isp = 900 s = u/g,
so u = (900 s)(9.8 m/s2) = 8,820 m/s
let v = Mach 15 = 15(343 m/s) = 5,145 m/s
mdot = (atm density)(intake area)(velocity) = rhoAv
at 25,000 m = 81,000 ft, density rho = 0.04 kg/m3
let intake diameter = 3 m, so A about 7 m2
mdot = (0.04 kg/m3)(7 m2)(5,145 m/s) = 1,441 kg/s
T = mdot(u - v) = (1,441 kg/s)(8,820 - 5,145) m/s = 5.3 MN
T= (5,300,000 N)(0.2248 lbf/N) = 1,190,000 lbf