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.