Suppose that $R$ is a commutative ring with identity $1$
Let $a\in R$ with $ab=0$ for some $b\ne0$.
Under what conditions $a$ must be also nilpotent?
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This is technically an answer, though I don't know if it's particularly useful...
Proof: If $a$ is nilpotent, then it is contained in all (minimal) prime ideals of $R$. Conversely, suppose that every minimal prime ideal containing $b$ also contains $a$. It is enough to show that $a$ is a member of every minimal prime of $R$ (for then $a$ is a member of every prime of $R$, and is therefore in the nilradical of $R$). So let $P$ be a minimal prime of $R$. If $b \in P$ then by hypothesis $a \in P$ also. On the other hand, if $b \notin P$ then $ab = 0 \in P$ with $P$ prime implies that $a \in P$. QED
In particular, if $b \in R$ is a zero divisor that is not a member of any minimal prime, then every element of $R$ that annihilates $b$ is nilpotent. (For if $ab = 0$, then it's vacuously true that every minimal prime containing $b$ also contains $a$.)
For a specific example of this situation, take the ring $R = k[x,y]/(x^2,xy)$. This has unique minimal prime ideal $(x)$, with $y \notin (x)$. So every element of $R$ that annihilates $y$ (for instance, $x$) must be nilpotent. (In fact, the annihilator of $y$ is quite easily seen to be $(x)$. So it's not a terribly interesting example.)
(You may also wish to see the following question on MathOverflow for vaguely related information: http://mathoverflow.net/questions/20826.)
I don't think there is any better answer for this question than the more specific ones given at your other question "Under what conditions does a ring R have the property that every zero divisor is a nilpotent element? ". The most natural conditions are mentioned there, and they are generally conditions on ideals of the ring, not the specific element $a$.
To recap some of the sufficient conditions mentioned there that make zero divisors nilpotent:
(Actually, $2$ implies $3$.)