= Solution
With $A^\mu=(0,\mathbf A)$, the lowered components are $A_\mu=(0,-\mathbf A)$. Define $\boldsymbol\pi=-i\boldsymbol\nabla+e\mathbf A$ and $\mathbf B=\boldsymbol\nabla\times\mathbf A$. This convention describes physical charge $q=-e$. Since $F_{ij}=-\epsilon_{ijk}B_k$,
$$
\sigma_{\mu\nu}F^{\mu\nu}=-2\begin{pmatrix}\boldsymbol\sigma\cdot\mathbf B&0\\0&\boldsymbol\sigma\cdot\mathbf B\end{pmatrix}.
$$
Writing $\psi=e^{-i(m+E)t}\phi$ makes the squared equation
$$
\left[(m+E)^2-m^2-\boldsymbol\pi^2-e\boldsymbol\Sigma\cdot\mathbf B\right]\phi=0.
$$
Drop $E^2$ relative to $2mE$ and write $\phi=(\varphi,\chi)^T$. Both two-component blocks obey the <Pauli equation>
$$
\boxed{E\varphi=\left[\frac{(\mathbf p+e\mathbf A)^2}{2m}+\frac e{2m}\boldsymbol\sigma\cdot\mathbf B\right]\varphi,\qquad
E\chi=\left[\frac{(\mathbf p+e\mathbf A)^2}{2m}+\frac e{2m}\boldsymbol\sigma\cdot\mathbf B\right]\chi.}
$$
The gauge-covariant kinetic term contains $\mathbf p^2/(2m)$, $e(\mathbf A\cdot\mathbf p+\mathbf p\cdot\mathbf A)/(2m)$, and $e^2A^2/(2m)$: free kinetic, orbital magnetic, and diamagnetic contributions. The spin term is $-\boldsymbol\mu\cdot\mathbf B$ with $\boldsymbol\mu=-e\boldsymbol\sigma/(2m)$, giving spin gyromagnetic factor $g=2$ at this order. In uniform $\mathbf B$ and symmetric gauge, the orbital term is $e\mathbf B\cdot\mathbf L/(2m)$.
Squaring introduces extra formal solutions unless the original first-order equation is retained. In the usual positive-energy Dirac basis it relates the small lower component to the upper one by $\chi=(\boldsymbol\sigma\cdot\boldsymbol\pi)\varphi/(2m+E)\simeq(\boldsymbol\sigma\cdot\boldsymbol\pi)\varphi/(2m)$. Thus these are not two independent nonrelativistic particles. Restoring units places $\hbar$ in the spin term, $e\hbar\boldsymbol\sigma\cdot\mathbf B/(2m)$.
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