Let $R$ be a commutative ring (with $1$) such that every non-zero divisor in $R$ is a unit (see Rings in which every non-unit is a zero divisor for various stabs at what these are called). Let $M$ be a finitely generated $R$-module.

My question:

Can we conclude that every non-zero divisor in $\mathrm{End}(M)$ is a unit?

**For example:** When $M$ is a free module, we have $\mathrm{End}(M) \cong R^{n \times n}$ is a matrix algebra and so every non-zero divisor is a unit (see Do these matrix rings have non-zero elements that are neither units nor zero divisors?).

A few notes:

- Even matrix algebras can misbehave when $R$ is non-commutative. Also, if $R$ has a non-zero divisor which is not a unit, then $\mathrm{End}(R)=R^{1 \times 1}=R$ has a non-zero divisor which is not a unit (thus the assumptions on $R$).
- If $V$ is an infinite dimensional vector space (over some field). Then $\mathrm{End}(V)$ has non-zero divisors which aren't units (thus the "finitely generated" assumption).
- If the answer to the question is "No" (I'd like a counter-example), then I would like to know under what circumstances $\mathrm{End}(M)$ does have this property. What assumptions on the module will force this to hold? (e.g. This holds when $M$ is a ??? module -- like flat or projective or something.) Or what assumptions on the ring will force this to hold for all modules? (e.g. This obviously holds when $R$ is a field.)
- Motivation and background: I've been working with a few undergraduates on a related problem. We tripped over this question and I have no idea whether this is true or not (I'm not a ring theorist and have a feeling this may be a difficult question to entirely resolve).

Thanks for reading my question! We (my group of undergrads and myself) appreciate your help!