= Solution
Keep only first order in the <metric perturbation> and use the <Minkowski metric> to raise and lower indices; set $G=c=1$. The <Levi-Civita connection> has
$$
\Gamma^{\rho(1)}{}_{\mu\nu}=\frac12\eta^{\rho\sigma}(\partial_\mu h_{\sigma\nu}+\partial_\nu h_{\sigma\mu}-\partial_\sigma h_{\mu\nu}).
$$
The quadratic products of <Christoffel symbols> in the <Riemann curvature tensor> can be dropped at this order. Contracting the remaining derivatives gives
$$
R^{(1)}_{\mu\nu}=\frac12\bigl(\partial_\rho\partial_\mu h^\rho{}_\nu+\partial_\rho\partial_\nu h^\rho{}_\mu-\Box h_{\mu\nu}-\partial_\mu\partial_\nu h\bigr),\qquad R^{(1)}=\partial_\mu\partial_\nu h^{\mu\nu}-\Box h,
$$
where $\Box=\eta^{\rho\sigma}\partial_\rho\partial_\sigma=-\partial_t^2+\nabla^2$ is the <d'Alembert operator>. In four dimensions the <trace-reversed metric perturbation> satisfies $\bar h=-h$ and $h_{\mu\nu}=\bar h_{\mu\nu}-\tfrac12\eta_{\mu\nu}\bar h$. Writing the <Linearized Einstein equations> in terms of it gives
$$
G^{(1)}_{\mu\nu}=\frac12\bigl(\partial_\rho\partial_\mu\bar h^\rho{}_\nu+\partial_\rho\partial_\nu\bar h^\rho{}_\mu-\Box\bar h_{\mu\nu}-\eta_{\mu\nu}\partial_\rho\partial_\sigma\bar h^{\rho\sigma}\bigr).
$$
In <Lorenz gauge in linearized gravity>, the first, second and fourth terms vanish. Therefore $G^{(1)}_{\mu\nu}=8\pi T_{\mu\nu}$ becomes
$$
\boxed{\Box\bar h_{\mu\nu}=-16\pi T_{\mu\nu}.}
$$
The divergence of this equation also gives $\partial^\mu T_{\mu\nu}=0$, the leading-order <stress-energy conservation> required by the <contracted Bianchi identity>.
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