= Solution
In the source's Gaussian electromagnetic units, set $\boldsymbol\Pi=-i\hbar\boldsymbol\nabla-e\mathbf A/c$ and take $\Phi=0$. Then
$$
\boxed{H_D=c\boldsymbol\alpha\cdot\boldsymbol\Pi+\beta mc^2=
\begin{pmatrix}mc^2I&c\boldsymbol\sigma\cdot\boldsymbol\Pi\\c\boldsymbol\sigma\cdot\boldsymbol\Pi&-mc^2I\end{pmatrix}.}
$$
The chosen gauge is independent of $x,z,t$, so simultaneous <energy> and canonical $p_x,p_z$ <Eigenstate>[eigenstates] justify the exponential ansatz. On the remaining spinors use $\Pi_x=p_x+eBy/c$, $\Pi_y=-i\hbar d/dy$, $\Pi_z=p_z$. The coupled equations are $(E-mc^2)\varphi=c\sigma\cdot\Pi\chi$ and $(E+mc^2)\chi=c\sigma\cdot\Pi\varphi$.
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