For which values of $n$ does $\mathbb{Z}/n\mathbb{Z}$ have the property that every non-zero element is either a unit or is nilpotent? Question: For which values of $n$ does $\mathbb{Z}/n\mathbb{Z}$ have the property that every non-zero element is either a unit or is nilpotent?
It is straightforward to see that for $n=p$ a prime, every non-zero element in $\mathbb{Z}/n\mathbb{Z}$ is unit. We can also see that every non-zero element of the ring $\mathbb{Z}/n\mathbb{Z}$ is either a unit or a zero-divisor.
However, I do not know how to extend this to the question above. I have identified this thread but still cannot figure out which other $n$ have the property described above. 
 A: A nonzero ring in which every element is either a unit or nilpotent is a local ring. As we know, $\mathbb{Z}/n$ is a local ring if and only if $n$ is a power of a prime.
Show that $\mathbb{Z}_n$ is local ring iff $n$ is a power of a prime number
A: Assume $n > 1$, and let $R=\mathbb{Z}/n\mathbb{Z}$.

For $x\in \mathbb{Z}$, let $\bar{x}$ denote the corresponding element of $R$.

Consider two cases . . .

Case $(1)$:$\;n$ is a power of a prime.

Let $n=p^k$, where $p$ is prime, and $k$ is a positive integer.

Suppose $\bar{x}$ is a non-unit of $R$.

Then we must have $p{\,\mid\,}x$.
\begin{align*}
\text{Then}\;\;&p{\,\mid\,}x\\[4pt]
\implies\;&p^k{\,\mid\,}x^k\\[4pt]
\implies\;&\overline{x^k}=0\\[4pt]
\implies\;&\bar{x}^k=0\\[4pt]
\end{align*}
hence $\bar{x}$ is a nilpotent element of $R$.

Case $(2)$:$\;n$ has at least two distinct prime factors.

Let $p,q$ be distinct prime factors of $n$.

Then $\bar{p}$ is not a unit of $R$, since $\gcd(p,n) = p > 1$.

Suppose $\bar{p}$ is a nilpotent element of $R$.

Then $\bar{p}^k=0$, for some positive integer $k$.
\begin{align*}
\text{Then}\;\;&\bar{p}^k=0\\[4pt]
\implies\;&\overline{p^k}=0\\[4pt]
\implies\;&n|p^k\\[4pt]
\implies\;&q|p^k\\[4pt]
\implies\;&q|p\\[4pt]
\end{align*}
contradiction.

Hence $\bar{p}$ is not a unit of $R$, and also not a nilpotent element of $R$.

Therefore, $R$ is such that every element is either a unit or nilpotent if and only if $n$ is a power of a prime.
