Can an infinite commutative ring have a finite (but nonzero) number of non-nilpotent zero-divisors? By a theorem of Ganesan, if a commutative ring not a domain has only finitely many zero-divisors, then the ring must be finite. (There are analogous results for non-commutative rings.)
There are plenty of examples of infinite rings with a finite number of nonzero nilpotents. There are also plenty of examples of infinite rings with an infinite number of zero-divisors, all of which are nilpotent.
However, I am unaware of any ring with an infinite number of zero-divisors, of which $0 < n < \infty$ are non-nilpotent.
Can anyone give an example or explain why this can't happen. I am mostly interested in the commutative case, but non-commutative examples would be interesting too.
 A: No, there is no such example.
Recall that the nilradical $N$ of $R$ is the ideal of nilpotent elements.  It equals the intersection of all prime ideals of $R$.
On the other hand, the set $D$ of zero-divisors of $R$ can be expressed as the union of the radicals of the annihilators of individual nonzero elements of $R$ (Atiyah-MacDonald Prop. 1.15):
$$D = \bigcup_{x\neq 0} \sqrt{(0:x)}$$
Here $(0:x)$ is an ideal, and its radical is the intersection of all the primes containing it.  Thus $D$ is a union of ideals, each of which contains the nilradical $N$.  If any of these ideals $I$ properly contains $N$, then if $N$ is infinite we conclude $I\setminus N$ is also infinite (since it contains a whole coset of $N$), and hence $D\setminus N$ is infinite.
EDIT:  Here's an easier proof in a different spirit, motivated by the preceding argument.
Suppose $x,y\in R$, such that $x$ is nilpotent and $y$ is a zerodivisor.  I claim $x+y$ is a zerodivisor.  Let $z\neq 0$ be such that $yz=0$.  If $xz=0$, we are done.  Otherwise, let $n$ be the smallest number such that $x^nz=0$ (which happens for some $n$ since $x$ is nilpotent).  Then $x^{n-1}z\neq 0$ but $x(x^{n-1}z)=0$, so $(x+y)x^{n-1}z=0$.  Thus $x+y$ is a zerodivisor.
Now if $y$ is not nilpotent, $x+y$ is not nilpotent since the nilradical $N$ is an ideal.  It follows that the coset $N+y$ consists entirely of nonnilpotent zerodivisors, so if $N$ is infinite then there are infinitely many nonnilpotent zerodivisors.
