Solution: Let $a$ be a nonzero element of $\mathbb{Z} _n$. Suppose that $a$ is a unit. We claim that $a$ can not be a zero divisor. If $a$ is a zero divisor, then there exists a nonzero $b \in \mathbb{Z} _n$ such that \[ a b = 0 \implies a^{-1} \left( a b \right) = 0 \implies b = 0, \] a contradiction. Thus, $a$ is not a zero divisor.

On the other hand, if $a$ is a zero divisor, then $a$ can not be a unit. If not, then there exists $c \in \mathbb{Z} _n$ such that $ac = 1$. Since $a$ is a zero divisor, there exists a nonzero element $d\in \mathbb{Z} _n$ such that $ad = 0$. We have \[ ad = 0 \implies c (ad) = 0 \implies d = 0, \] a contradiction. Thus, $a$ can not be a unit.