It is a standard theorem that the nilradical of a ring $R$ is the intersection of all the prime ideals in $R$.
So, for example, take $\mathbb{Z}/12$. Its prime ideals are (the images of) $(3)$ and $(2)$ (using the correspondence that prime ideals in a quotient ring $R/I$ are precisely the prime ideals in $R$ containing $I$). The intersection of $(3)$ and $(2)$ is the ideal $(6)$, which contains the two elements $0$ and $6$.
Edit: If you don't want to use the quoted theorem, then, at least in PID's, an element $x$ in $R/(r)$ is nilpotent if and only if $r|x^n$ for some $n$ (this is easy to prove!). So for the example $\mathbb{Z}/12$, we see that $12|6^2$, så $6^2=0$ is nilpotent.
Note that the condition $r|x^n$ for some $n$ is equivalent that all factors of $r$ must divide some factor of $x$.