There's a very famous thought experiment that verifies the temperature dependence of blackbody spectra
The proof first assumes that two blackbodies have different spectra and then shows that this leads to a contradiction. Let two blackbodies A & B, both at temperature $T$, radiate different spectra. Then use a filter and aperture stops to allow them to transfer heat only by radiation in a given passband. Then the radiation of A is entirely absorbed by B, and the radiation of B is entirely absorbed by A. Thus if their spectra are different, there would be a net transfer of heat between A & B, but their temperatures are the same. Since heat transfer between objects of the same temperature does not occur, the spectra must be identical. The choice of filter passband was arbitrary, so the spectra must be identical at all frequencies.
I understand it this way: if A released photons of a single wavelength, so should B, otherwise there will be a net energy transfer (photons have no way to go apart from the other body). But consider this – A emits photons of energy $2E$, while B emits photons of energy $3E$ and $E$ in equal amounts. In this case they would be exchanging equal energy (so the temperatures would remain same), but the spectra would be different.
I know that this is perhaps impossible in real life, but isn't this a case that clearly is not explained by the thought experiment?
Best Answer
Your proposal works in the absence of the filter passbands, but the main point of the argument is the use of filter passbands. Ij your particular case, the counterexample breaks when you pick, for example, a filter passband sharply centered around the frequency $\frac{2E}{\hbar}$. In this case, no photons of $B$ will be able to cross the filter passband, while all photons of $A$ will cross. Hence, $A$ will cool down while $B$ heats up, and hence they can't be in thermal equilibrium, which is a contradiction.
Notice the importance of the filter passband in the argument. That is what allows the selection of frequencies and permits the argument to work for each individual frequency.