Here is a proof which may not be the best but demonstrates some standard techniques:

Since $R$ has finite inj. dim. one can replace $M$ by a high syzygy, so one can assume $M$ has full depth. Thus one can kill a full regular sequence for both $M$ and $R$ and (as finiteness of proj. or inj. dim are not affected) assume $R$ is Artinian. Now, your last question shows that you already know in this case $M$ must be injective.

Map a free module onto $M$ and look at the exact sequence:

$$ 0 \to N \to R^n \to M \to 0$$

As $R$ is Gorenstein, $N$ also has fin. inj. dim., so injective. But then $\text{Ext}_R^1(M,N)=0$, hence the sequence splits, and $M$ must be free.

PS: The name is Bruns. You said in your profile that you are a graduate student interested in commutative algebra. If that is indeed the case, then perhaps taking a serious course and talking to the experts at your institution would be more effective then learning it on MO (-: Good luck!

Here is a proof which may not be the best but demonstrates some standard techniques:

Since $R$ has finite inj. dim. one can replace $M$ by a high syzygy, so one can assume $M$ has full depth. Thus one can kill a full regular sequence for both $M$ and $R$ and (as finiteness of proj. or inj. dim are not affected) assume $R$ is Artinian. Now, your last question shows that you already know in this case $M$ must be injective.

Map a free module onto $M$ and look at the exact sequence:

$$ 0 \to N \to R^n \to M \to 0$$

As $R$ is Gorenstein, $N$ also has fin. inj. dim., so injective. But then $\text{Ext}_R^1(M,N)=0$, hence the sequence splits, and $M$ must be free.

PS: The name is Bruns. You said in your profile that you are a graduate student interested in commutative algebra. If that is indeed the case, then perhaps taking a serious course and talking to the experts at your institution would be more effective then learning it on MO (-: Good luck!