Matt Samuel already gave a good answer with both an example and some thoughts on why this should be rare. This answer intends to add a different perspective on the proportion of the rings with all zero divisors nilpotent. $\newcommand\AA{\Bbb{A}}\newcommand\PP{\mathbb{P}}\newcommand\Disc{\operatorname{Disc}}$
One perspective on what proportion of rings have no nonnilpotent zero divisors is the following. Note that this is really rough first thought.
First let's be careful though, rings with no zero-divisors trivially have this property, so I'll try to consider what rings with zero-divisors have the property that all zero-divisors are nilpotent.
Step 1: Set up a space to parametrize a nice class of rings
Let $k$ be an algebraically closed field. Let $n,d_1,d_2 > 0$. Let $x=(x_1,x_2,\ldots,x_n)$ be coordinates on $\AA^n_k$. Let $r_1=\binom{n+d_1}{d_2}$, which is the number of monomials in $k[x_1,\ldots,x_n]$ of degree at most $d_1$. Thus there are $r_1$ multiindices $I$ of degree at most $d_1$. Let $r_2=\binom{n+d_2}{d_2}$ as well. Let $c=(c_I)$ be coordinates on $\PP^{r_1-1}_k$, and let $d=(d_J)$ be coordinates on $\PP^{r_2-1}_k$ as $I$ ranges over multiindices of degree at most $d_1$ and $J$ ranges over multiindices of degree at most $d_2$.
Then consider the subvariety, $V$, of $\AA^n_k\times_k\PP^{r_1-1}_k\times_k \PP^{r_2-1}_k$ cut out
by the polynomial $$f_{c,d}(x)=g_c(x)g_d(x):=\left(\sum_I c_Ix^I\right)\left(\sum_J d_Jx^J\right).$$
$V$ is equipped with a natural map $V\to \PP^{r_1-1}_k\times_k\PP^{r_2-1}$, and the fiber over a point $(a,b)$ in $\PP^{r_1-1}_k\times_k\PP^{r_2-1}_k$ is the subvariety of $\AA^n_k$ cut out by the degree at most $d_1+d_2$ reducible polynomial $$f_{a,b}(x)=g_a(x)g_b(x).$$
This subvariety of $\AA^n_k$ can be thought of as corresponding to the ring $k[x_1,\ldots,x_n]/(g_ag_b)$, so we can think of our variety $V$ as parametrizing a certain nice class of rings all (except for a Zariski closed subset where $g_a=0$ or $g_b=0$) of which are guaranteed to have zero-divisors, since $g_a$ and $g_b$ are zero divisors.
Step 2: Investigate the points corresponding to rings with the desired property and related properties
We can then ask if we can characterize the subset of $V$ corresponding to rings in which all zero-divisors are nilpotent.
Well, what are the zero-divisors in $k[x_1,\ldots,x_n]/(f_{a,b})$? Since $k[x_1,\ldots,x_n]$ is a UFD, handily enough, the zero-divisors are precisely the factors of $f_{a,b}$. Moreover $k[x_1,\ldots,x_n]/(f_{a,b})$ has nilpotents if and only if $f_{a,b}$ is not square free (take the radical of $f_{a,b}$, it will be nilpotent if and only if it is nonzero, if and only if $f_{a,b}$ is not square free). However the ring $k[x_1,\ldots,x_n]/(f_{a,b})$ satisfies our property that all zero-divisors are nilpotent if and only if $f_{a,b}$ is a power of an irreducible polynomial.
Step 3: Conclude
Note that if $f_{a,b}$ is square-free if and only if it is relatively prime to $f_{a,b}'$, which is true if and only if $\Disc(f_{a,b})\ne 0$. Thus every point $(a,b)\in\PP^{r_1-1}_k\times_k\PP^{r_2-1}_k$ corresponding to a ring with any nilpotents at all, let alone one in which every zero-divisor is nilpotent, satisfies a polynomial equation, $\Disc(f_{a,b})=0$. Thus rings in our parametrized class with any nilpotents at all correspond to a (proper) Zariski closed subset of $\PP^{r_1-1}_k\times_k\PP^{r_2-1}_k$, which is irreducible (see here).
Thus rings in our parametrized class satisfying your property (all zero-divisors are nilpotent) are contained in a codimension 1 class of rings (any nilpotents). Hence almost all rings in our parametrized class have zero-divisors that are not nilpotent.
Notes
This is really rough. The parametrized class of rings is a very specific, very nice subset of $k$-algebras. Nonetheless, hopefully it will give you (or other readers) intuition on why very few rings should have the property you want (as long as they have some zero divisors of course).
I needed to eliminate integral domains because I chose to parametrize hypersurfaces, and most hypersurfaces are irreducible. (Check out the link, Qiaochu Yuan gives a really nice quick proof of this fact). There's a reasonable chance that I wouldn't have needed to eliminate integral domains if I'd chosen e.g. codimension 2 subvarieties, but those are much harder to characterize.