= Solution
In <Feynman gauge>, the Lagrangian differs by a total divergence from
$$
\mathcal L=-\frac12\partial_\mu A_\nu\partial^\mu A^\nu.
$$
Its <Euler-Lagrange equation> is $\Box A^\mu=0$. Directly from the original gauge-fixed Lagrangian, the canonical momenta are
$$
\pi^\mu=-F^{0\mu}-\eta^{0\mu}\partial_\nu A^\nu,
$$
equivalently $\pi^\mu=-\partial^0A^\mu$ after using the total-divergence form.
Using the mode expansions and oscillator commutator, the two nonzero cross terms combine to
$$
\begin{aligned}
[\widehat A_\mu(\mathbf x),\widehat\pi_\nu(\mathbf y)]
&=\frac{i}{2}\int\frac{d^3p}{(2\pi)^3}
\sum_\lambda\eta_{\lambda\lambda}
\epsilon_\mu^\lambda(\mathbf p)\epsilon_\nu^\lambda(\mathbf p)
\left[e^{i\mathbf p\cdot(\mathbf x-\mathbf y)}
+e^{-i\mathbf p\cdot(\mathbf x-\mathbf y)}\right]\\
&=i\eta_{\mu\nu}\delta^{(3)}(\mathbf x-\mathbf y),
\end{aligned}
$$
where polarization completeness was used. The other equal-time field commutators vanish.
Substitution into the Hamiltonian and <normal ordering> give
$$
:\widehat H:
=-\sum_{\lambda=0}^3\eta_{\lambda\lambda}
\int\frac{d^3p}{(2\pi)^3}|\mathbf p|\,
\widehat a_{\mathbf p}^{\lambda\dagger}\widehat a_{\mathbf p}^{\lambda}.
$$
Every oscillator excitation carries positive energy $|\mathbf p|$, but the covariant state space has an indefinite inner product: time-like excitations have negative norm, and time-like and longitudinal polarizations are unphysical. <Gupta-Bleuler quantization> imposes
$$
(\partial_\mu\widehat A^\mu)^{(+)}|\mathrm{phys}\rangle=0
$$
and quotients by null states. The physical state space contains only the two transverse photon polarizations with positive norm, as required by <gauge invariance>.
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