= Solution
In SI units, <Maxwell's equations> are
$$
\nabla\cdot\mathbf E=\rho_e/\epsilon_0,\quad\nabla\cdot\mathbf B=0,\quad
\nabla\times\mathbf E=-\partial_t\mathbf B,\quad
\nabla\times\mathbf B=\mu_0\mathbf J+\mu_0\epsilon_0\partial_t\mathbf E.
$$
Dotting the last equation with $\mathbf E/\mu_0$, and using the curl equation for $\mathbf E$, gives
$$
\partial_t\left(\frac{\epsilon_0}2|\mathbf E|^2+\frac1{2\mu_0}|\mathbf B|^2\right)
=-\mathbf E\cdot\mathbf J-\frac1{\mu_0}\nabla\cdot(\mathbf E\times\mathbf B).
$$
Integrate over the fixed region and apply the <divergence theorem>. This proves <Poynting's theorem>: the rate of increase of electromagnetic <energy> stored inside, plus the outward <Poynting flux>, plus the rate of work on charges, is zero. The last term need not be positive for arbitrary externally driven currents; when positive it is power transferred from the <field> to matter.
For the travelling wave put $\varphi=kz-\omega t$. A purely oscillatory <magnetic field> satisfying <Faraday's law> is
$$
\mathbf B=(0,(k\alpha/\omega)\cos\varphi,0).
$$
Both divergences vanish. The <vacuum> <Ampère-Maxwell equation>[Ampère-Maxwell law] then requires $k^2=\mu_0\epsilon_0\omega^2$, so with the positive constants in the question,
$$
\boxed{\omega=ck,\quad c=(\mu_0\epsilon_0)^{-1/2},\quad\mathbf B=(0,(\alpha/c)\cos\varphi,0)}.
$$
A superposed static background is not part of the pure plane-wave solution. The electric and magnetic <energy> contributions are equal, giving
$$
\mathcal U=\epsilon_0\alpha^2\cos^2\varphi,\qquad
\mathbf P=\frac{\alpha^2}{\mu_0c}\cos^2\varphi\,\mathbf e_z
=c\mathcal U\,\mathbf e_z.
$$
Consequently $\boxed{|\mathbf P|=c\mathcal U}$ pointwise, including the nodes where both sides vanish.
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