There is an $n^k$ prover if and only if $P = NP$
An $n^k$ prover implies that the set $\{\left(\varphi,1^n\right)|\text{$\varphi$ has a proof of length $\le n$}\}$ is in $P$, since we can simply run the prover for $n^k$ steps and check whether it yields a valid proof.
It remains to show that this language is NP complete. This follows from a straightforward reduction from SAT. Given a CNF $\varphi$ over the variables $x_1,...,x_n$ return the pair $\left(\exists x_1,...,x_n: \varphi, 1^k\right)$ where $k=k(|\varphi|)$ is some upper bound on the length of the satisfiability proof for satisfiable formulas of length $|\varphi|$.