= Solution
Use $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$. The stated derivative convention gives $[D_\mu,D_\nu]=-ieF_{\mu\nu}$. Also $[\gamma^\mu,\gamma^\nu]=-2i\sigma^{\mu\nu}$. Splitting products into symmetric and antisymmetric parts therefore yields
$$
(\gamma^\mu D_\mu)^2=D_\mu D^\mu+\frac14[\gamma^\mu,\gamma^\nu][D_\mu,D_\nu]
=D\cdot D-\frac e2\sigma^{\mu\nu}F_{\mu\nu}.
$$
Multiplying the first-order equation by $i\gamma^\mu D_\mu+m$ now gives
$$
\boxed{\left[-D\cdot D+\frac e2\sigma_{\mu\nu}F^{\mu\nu}-m^2\right]\psi=0.}
$$
The antisymmetric derivative commutator is the electromagnetic field strength. It is the extra term absent when squaring an uncoupled Dirac operator with commuting derivatives, and it generates the spin–magnetic-field interaction in the nonrelativistic limit.
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