We cannot use that the radiated power of each body is simply determined by the 4th power of its temperature. That result only holds if we have black body radiation, which depends on frequency $\nu$ as: (See: [1] )
$$
B_\nu(T) = \frac{2h\nu^3}{c^2}\ \frac{1}{e^{h\nu/kT}-1}
$$
and integrating this over frequencies (see [2]) does indeed give a result $\sim T^4$. Non-black bodies have some, in general frequency dependent, emissivity, $\varepsilon_\nu$, which changes the radiated power to:
$$
P_\nu(T) = \varepsilon_\nu\ B_\nu(T)
$$
and the integral over frequency will in general no longer be $C\ T^4$ for any constant $C$, because for different $T$-values, different frequency ranges become dominant, so the frequency dependence introduced by $\varepsilon_\nu$ can change the $T$-dependence.
What we do know, however, is that the absorptivity is the same as the
emissivity. We now look at the power balance, by computing the power received by body 2 from body 1, and vice versa, $P_{21}$ and $P_{12}$:
$$
P_{21} = \int\limits_0^\infty d\nu \ \varepsilon_{2,\nu}\ \varepsilon_{1,\nu}\ B_\nu(T_1) \\[5pt]
P_{12} = \int\limits_0^\infty d\nu \ \varepsilon_{1,\nu}\ \varepsilon_{2,\nu}\ B_\nu(T_2)
$$
It is obvious that if $T_1=T_2$ then $P_{21}=P_{12}$. But the converse is not immediately clear, so the question remains: can there be another equilibrium point? Let's assume, without loss of generality, that we have such a solution with $T_2>T_1$. Let's observe that
$ \varepsilon_{1,\nu}$ and $ \varepsilon_{2,\nu} $ are both non-negative functions, and then use:
$$\begin{align}
T_2-T_1>0 \ &\Rightarrow B_\nu(T_2) - B_\nu(T_1) >0\ \ \ \text{for all finite}\ \nu, \\[5pt]
\Rightarrow P_{12}-P_{21}&=\int\limits_0^\infty d\nu \ \varepsilon_{1,\nu}\ \varepsilon_{2,\nu}\,\big(B_\nu(T_2)-B_\nu(T_1)\big) >0,
\end{align}$$
unless $\varepsilon_{1,\nu}\ \varepsilon_{2,\nu}=0$ everywhere. So there can only exist an equilibrium $P_{12}=P_{21}$ with $T_1\neq T_2$ if:
- For one of the bodies $\varepsilon_\nu=0$ at all frequencies, so we have a perfectly white body,
- Or at all frequencies either $\varepsilon_{1,\nu}=0$ or $\varepsilon_{2,\nu}=0$, so none of the bodies is perfectly white but they both have perfectly nonabsorbing frequency regions, which combine to fill the whole frequency range.
It should be added that stricly speaking we would have to integrate over all angles of emission and all points on the surface of each body, but for the sake of this proof we can assume that those integrals have for each frequency $\nu$ been done and are absorbed into the definition of $\varepsilon_{1,\nu}$ and $\varepsilon_{2,\nu}$.
