Under what conditions does a ring R have the property that every zero divisor is a nilpotent element? Under what conditions does a ring $R$ have the property that every zero divisor is a nilpotent element ? 
If we have a ring $R$, we know that every nilpotent element is either zero or a zero divisor.
If $A$ is the set of all nilpotents except the zero, and $B$ is the set of all nonzero zero divisors, then under what conditions on $R$ do we have $A = B$ ? 
 A: Most generally speaking, a nontrivial idempotent is a zero divisor that is never nilpotent. From this we deduce that such a ring cannot have any nontrivial idempotents.
The exclusion of nontrivial idempotents is really strong. In particular, it implies that none of the nontrivial right or left ideals are summands. (Commutative domains are a case of this, of course, but in that case the zero divisors and nilpotent elements are all 0 :) ) 
Here is a crisp partial result. In the case when $R$ is right Artinian, all elements are either units or zero divisors. A right Artinian ring without nontrivial idempotents is local. Since the maximal ideal is nilpotent, it's clear that all the zero divisors are nilpotent. In summary, this says that among right Artinian rings, the local ones are exactly the ones with the "ZD implies nilpotent" property. 
I've seen Jacoson call commutative rings without nontrivial idempotents connected rings, but I have to say that I don't have many exotic examples of them in mind. I'm guessing that being connected probably doesn't characterize the "ZD implies nilpotent" property.
There are a lot of examples of reduced rings which aren't domains which are non-examples for your property. For example, $F\times F$ for a field $F$ has no nilpotent elements but obviously has zero divisors. This ring is already pretty 'nice' (commutative Artinian!) and you can even make it finite by using a finite field.

D. Lazard pointed out to me in a conversation that $\{0\}$ is a primary ideal iff $Nil(R)$ is prime and there is no other associated prime. This is very cool too :)
A: If $A$ is a commutative ring with this property, then the zero-divisors form an ideal  of $A$ (since the nilradical is an ideal), in fact a prime ideal, and thus it's the unique minimal prime ideal of $A$.  Conversely, if $A$ has a unique minimal prime ideal containing all the zero-divisors, then this has to be the nilradical, so every zero-divisor is nilpotent.
People seem to be calling these rings 'primary rings,' since this condition characterizes the rings you get after quotienting an arbitrary ring by a primary ideal.  I don't know how widely used this term is.
