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.
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.