Let $A$ be your ring and $X=\mathrm{Spec} A$. The minimal primes of $A$ correspond to the irreducible components of $X$. An element of $f\in A$ induces a function $\widehat f:X\to \coprod_{\mathfrak p\in \mathrm{Spec} A}\kappa(\mathfrak p)$ and this function vanishes everywhere if and only if $f\in\mathfrak p$ for all $\mathfrak p\in \mathrm{Spec} A$, that is, when it is nilpotent. For an $f\in A$ contained in a minimal prime the induced function $\widehat f$ vanishes on the irreducible component corresponding to the minimal prime containing $f$.
If $X$ has a single irreducible component, then the functions induced by the elements of $A$ in the corresponding single prime ideal are vanishing everywhere hence they are nilpotent (in particular zero-divisors).
If $X$ has more than one irreducible component, then taking one element in each minimal prime, their product will produce a function that is vanishing everywhere and hence nilpotent.
Comment Apparently this is essentially the same proof as the one Graham included in the comments, but I can't let go any chance of giving a geometric proof of an algebra question. It appears longer, but only because i) setup ii) more details. Also, let me add that the length is the main reason I did not include it as a comment. In any case algebra=geometry. :)