= Solution
For a massive particle parametrized by <proper time>, the <geodesic equation> is
$$
\frac{d^2x^\mu}{d\tau^2}+\Gamma^\mu{}_{\alpha\beta}\frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau}=0,
\qquad g_{\alpha\beta}\frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau}=-c^2.
$$
The <weak-field approximation> alone does not imply Newtonian motion. Also take a <nonrelativistic limit>, $|v|/c\ll1$, in the usual quasistatic regime so the potentials' time variation does not compensate for the small velocity factors. Reparametrizing by coordinate time and retaining the leading velocity-independent spatial term gives
$$
\boxed{\frac{d^2x^i}{dt^2}=-c^2\Gamma^i{}_{00}=-\partial_i\Phi.}
$$
In empty space the first field equation gives $\nabla^2\Psi=0$. For static potentials, the spatial equations imply $\nabla^2(\Phi-\Psi)=0$ and then $\partial_i\partial_j(\Phi-\Psi)=0$. The difference is at most affine; isolated-system boundary conditions fixing both potentials to zero at infinity remove it. Thus $\Phi=\Psi$ and
$$
\boxed{\nabla^2\Phi=0\quad\text{in vacuum},}
$$
the empty-space <Newton-Poisson equation>. With the same quasistatic assumptions and dust present it becomes $\nabla^2\Phi=4\pi G\rho$. Keeping rapid time dependence instead leaves $\nabla^2\Phi=-3\partial_0^2\Psi$ in vacuum, so the static qualification matters.
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