Let $A$ be a Gorenstein noetherian local ring, and let $M$ be an $A$-module of finite injective dimension.
If $M$ is a finite $A$-module, it is easy to show these assumptions imply that $M$ has finite flat dimension (just use the fact that $A$ is both perfect and a dualizing complex).
What if $M$ is not finitely generated? is it still true that finite injective dimension implies finite flat dimension?
Suppose that $A$ is a not necessarily commutative ring, which is both left and right Noetherian. Given a left $A$-module $N$, and a right $A$-module $M$, there is a general (fourth quadrant, cohomological) spectral sequence due to Ischebeck, with $E_2$ term
$$E_2^{p,-q} = \operatorname{Ext}^p_A(\operatorname{Ext}^q_A(N,A),M)$$
which converges to
$$\operatorname{Tor}^A_{q-p}(M,N)$$
whenever $M$ has finite injective dimension. If, in addition, $A$ has finite injective dimension $d = \operatorname{id}(A) < \infty$, then whenever $n = q-p$ with $0 \leq q \leq d$ and $p \geq 0$, we must have $n \leq d$. Thus $\operatorname{Tor}^A_n(M,N)$ vanishes for any left $A$-module $N$ and any $n > d$. So, if $M$ has finite injective dimension, then
$$\operatorname{fd}(M) \leq \operatorname{id}(A)$$
which gives a positive answer to your question because your ring $A$ is assumed to be Gorenstein, hence has finite self-injective dimension.
You can find an English translation of Ischebeck's paper here; see Theorem (1.8).
