= Solution
Choose the <vector potential> $A=(-\int_0^yB(s)ds,0,0)$, whose curl is $(0,0,B(y))$, and a spin-down wavefunction $\Psi=e^{-ikx}\chi(y)(0,1)^{\mathsf T}$, independent of $z$. Define
$$
a(y)=\frac e\hbar\int_0^yB(s)ds,\qquad b(y)=\frac{eB(y)}\hbar,\qquad\epsilon=\frac{2mE}{\hbar^2}.
$$
The covariant $x$ derivative acts as $-i(k+a)$, and $\sigma_3$ acts as $-1$. The <Pauli equation> consequently reduces to
$$
-\chi''+(k+a)^2\chi-b\chi=\epsilon\chi,\qquad\boxed{a'=b>eB_0/\hbar>0}.
$$
The sign of the plane-wave phase was chosen to match the printed $k+a$ convention; choosing $e^{ikx}$ simply reverses the label $k$.
Set $W=k+a$ and $M=\partial_y+W$. On the square-integrable operator domain with vanishing boundary terms, $M^\dagger=-\partial_y+W$, and
$$
M^\dagger M=-\partial_y^2+W^2-W'=-\partial_y^2+(k+a)^2-b.
$$
Taking the inner product with $\chi$ gives $\epsilon\|\chi\|_2^2=\|M\chi\|_2^2\ge0$. Thus $\boxed{E\ge0}$.
A zero mode solves $M\chi=0$, so
$$
\boxed{\chi_k(y)=C_k\exp\!\left[-ky-\int_0^y a(s)ds\right].}
$$
Since $a(0)=0$ and $a'\ge b_0=eB_0/\hbar$, integration on either side of zero gives $\int_0^y a(s)ds\ge b_0y^2/2$. Therefore $|\chi_k(y)|\le|C_k|e^{-ky-b_0y^2/2}$, proving square integrability for every real $k$. Each can be normalized and supplies a distinct momentum label at zero energy. Hence \b[zero energy is infinitely degenerate]. In an infinite $x$ direction the individual plane waves are generalized momentum states; square-integrable superpositions of their normalized transverse zero modes give ordinary transverse states by the Fourier isometry. The stipulated zero $z$ momentum is understood per unit longitudinal length, or with periodic longitudinal normalization.
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