= Solution
Put $a_h^2=k^2+m^2>0$ and $\Lambda=a_h^2+\pi^2$. The plane-layer <poloidal-toroidal decomposition> supplies the assumed poloidal fields as
$$
\mathbf u=\nabla\times\nabla\times(\phi\hat{\mathbf z}),\qquad
\mathbf b=\nabla\times\nabla\times(\chi\hat{\mathbf z}).
$$
Since $w=-\Delta_h\phi$ and $b_z=-\Delta_h\chi$, potentials proportional to $e^{st+ikx+imy}\sin\pi z$ have the required vertical components. Their horizontal components are proportional to $\cos\pi z$, satisfying the <stress-free boundary conditions> and the horizontal-field <boundary conditions>. The double-<curl> representation automatically enforces <incompressibility> and the <solenoidal magnetic-field constraint>.
Let $W,\Theta,C$ be the amplitudes of $w,\theta,b_z$. The temperature and <resistive induction equations> immediately give
$$
(s+\Lambda)\Theta=W,\qquad (s+\zeta\Lambda)C=imW.
$$
Eliminate the total-pressure perturbation by applying $\Delta$ to the vertical momentum equation and subtracting $\partial_z$ times its <divergence>. The resulting vertical equation is
$$
\Lambda(s/\sigma+\Lambda)W=Ra_h^2\Theta+\zeta Q\,im\Lambda C.
$$
In particular, $imC=-m^2W/(s+\zeta\Lambda)$ shows explicitly that <magnetic tension> opposes the displacement.
Taking the <determinant> of these three amplitude equations gives the <horizontal-field magnetoconvection dispersion relation>:
$$
\boxed{\Lambda(s+\Lambda)\left[(s+\sigma\Lambda)(s+\zeta\Lambda)+\sigma\zeta Qm^2\right]
-\sigma Ra_h^2(s+\zeta\Lambda)=0.}
$$
This cubic <polynomial> includes all roots of the assumed poloidal coupled system; using its determinant form avoids discarding possible roots by dividing by $s+\Lambda$ or $s+\zeta\Lambda$. As a check, $Qm^2=0$ factors off magnetic diffusion and leaves the standard stress-free <Rayleigh-Bénard convection> growth equation.
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