In this question a user wondered if all finitely generated modules over a Dedekind finite ring (a ring satisfying $xy=1\implies yx=1$ for all $x,y$) are Hopfian (every surjective homomorphism $M\to M$ is injective.)
As pointed out in the question, it is possible that there exists an $n$ such that $R^n$ is not Hopfian, so there are in fact counterexamples.
So at this point, we see it is necessary for $R$ to be stably finite (that is, all of its matrix rings are Dedekind finite.)
Is stable finiteness sufficient to make all finitely generated $R$ modules Hopfian? Or is it perhaps equivalent to a stronger condition on $R$?
I suspect the answer is known, but I didn't spot an argument in the short while I've been thinking about it.
It's pretty obvious that $R$ being right Noetherian is sufficient, and this condition is strictly stronger than being stably finite. Another sufficient condition is having stable range 1.