It seems to me that the answer is NO. Namely, if $(M,g)$ is compact, conformally flat and has zero scalar curvature then $g$ is a flat metric. Indeed, by the hypothesis there is a smooth function $f:M \to \mathbb R$ such that $e^f g$ is a flat metric on $M$. Then a finite covering of $M$ is a torus. The metric $g$ then lift to such a torus and by the Gromov-Lawson theorem (see Corollary A in http://www.ihes.fr/~gromov/PDF/8%5B26%5D.pdf) the lifted metric is flat hence the original $g$ is flat. QED
As Robert Bryant observed I made a mistake assuming that $e^f g$ is a flat metric on $M$. Indeed, conformally flat means that $e^f g$ has constant sectional curvatures so the sectional curvature can be also 1 or -1. Watching the formula for the scalar curvature of a conformal change
https://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry
it seems that the problem reduce to the question if the number $(n-2)/4(n-1)$ is an eigenvalue of the Laplacian of a compact hyperbolic manifold of dimension $n$.
The same formula allows to rule out the flat case without using Gromov-Lawson theorem. Indeed, if $e^f g$ is flat and $g$ has zero scalar curvature then $e^{n-2}f$ is harmonic hence constant since $M$ is compact. So $g$ is flat.