= Solution
Use $G=c=1$ and the physical <real part> of the printed <plane wave>. Its null wavevector obeys $k^\alpha k_\alpha=0$, so each component satisfies $\Box h_{\alpha\beta}=0$. Direct differentiation of the defined stress tensor gives
$$
\partial^\nu\tau_{\mu\nu}=\frac1{32\pi}(\partial^\nu\partial_\mu h^{\alpha\beta})(\partial_\nu h_{\alpha\beta})
+\frac1{32\pi}(\partial_\mu h^{\alpha\beta})\Box h_{\alpha\beta}
=\frac12\partial_\mu\tau^\nu{}_\nu.
$$
The <trace> is proportional to $k^\nu k_\nu$ for a single <plane wave> and hence vanishes. Therefore
$$
\boxed{\tau^\nu{}_\nu=0,\qquad\partial^\nu\tau_{\mu\nu}=0.}
$$
This is local <conservation of energy> and <momentum conservation>; spatial integration gives conserved total charges when the boundary flux vanishes. For a finite-amplitude complex phasor, one takes the real field before quadratic products and may time-average: $\langle\tau_{\mu\nu}\rangle=k_\mu k_\nu H^{\alpha\beta}\overline{H}_{\alpha\beta}/(64\pi)$ in <transverse-traceless gauge>. Its <energy density> is nonnegative and its flux travels in the positive $z$ direction.
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