Does it follow from "for all finitely generated $N$, there exists $n_0$ such that for $n\geq n_0, \mathrm{Ext}^n(M,N)=0$" that $M$ has finite projective dimension, for a finitely generated $M$ ?
What if the ring $R$ is a finitely generated $k$-algebra, for a certain field $k$ ? If it's $kG$, for a finite group $G$ ? If it's $kG$, and I'm also assuming $\mathrm{Ext}^n(N,M)=0$ for $n$ large enough (how large still depending on $N$ though) ?
Of course, if $n_0$ doesn't depend on $N$ it's clearly the case; but I saw some paper which seems to claim that more general statement in the case of $kG$.
I know how to prove it when $G$ is a $p$-group and $\mathrm{char} k = p$, using a minimal projective resolution and the fact that any projective has a nonzero map to the simple module $k$; but I'm not sure that's true for more general $G$; and the paper is not very clear about whether $G$ is assumed to be a $p$-group or not, so I'd like to know how general the result is.
This is true whenever $R$ is Artinian. Indeed, in that case you can take $N$ to be a direct sum of all the simple modules up to isomorphism (there are only finitely many since all of them must appear in a composition series of $R$), and use the same argument as you used in the case that $G$ is a $p$-group. Explicitly, take a minimal resolution of $M$ (projective covers exist since everything is Artinian). Each map in this resolution has superfluous image, and thus vanishes when composed with any map to a simple module. This means all the maps in the resolution become $0$ after Homming into $N$. But each nontrivial term of the resolution has a nontrivial map to some simple module, and thus every term remains nonzero after Homming into $N$. We thus find that $\operatorname{Ext}^n(M,N)$ is nontrivial for all $n$ unless the resolution terminates, i.e. unless $M$ has finite projective dimension.
I believe it's also true when $R$ is Noetherian and commutative: if $R$ is local, the argument above still works since there is only one simple module and projective covers exist. So, $M$ has finite projective dimension after localizing at any prime. Then, since $M$ is finitely generated, if it has finite projective dimension locally at some prime, it has finite projective dimension in an entire Zariski neighborhood of that prime. We can then cover $\operatorname{Spec} R$ by finitely many of these Zariski neighborhoods to conclude $M$ has finite projective dimension globally.
I doubt it's true over an arbitrary ring, but don't know a counterexample.