= Solution
The <Lorenz gauge in linearized gravity> gives
$$
0=\partial^\mu\bar h_{\mu\nu}
=ik^\mu H_{\mu\nu}e^{ik\cdot x},
$$
so the symmetric amplitude is transverse:
$$
\boxed{k^\mu H_{\mu\nu}=0}.
$$
Under this condition the supplied linearized Einstein tensor reduces to
$$
G_{\mu\nu}=-\frac12\Box\bar h_{\mu\nu}.
$$
The vacuum <Einstein field equations> and the plane-wave ansatz then imply
$$
\boxed{k^\rho k_\rho=0}.
$$
\b[Thus the wavevector is null and the perturbation is a <plane gravitational wave in linearized gravity>. Lorenz gauge alone does not require the trace of $H_{\mu\nu}$ to vanish; that is a further gauge choice.]
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