Describe the zero divisors of $\mathbb{Z}/n\mathbb{Z}$ Am I correct in reasoning that all non-zero elements of $\mathbb{Z}/n\mathbb{Z}$ that have order smaller than $n$ are zero divisors? Since all residue classes that are smaller than $\overline{n}$ are in $\mathbb{Z}/n\mathbb{Z}$ we can find some $\overline{b} \in \mathbb{Z}/n\mathbb{Z}$ for every element $\overline{a} \in \mathbb{Z}/n\mathbb{Z}$ (with $|\overline{a}|<n$) so that $\overline{a} \overline{b}=\overline{0}$.
Also, is there some well known theorem stating this? 

Yes, the statement is
$\mathbb Z_n$ with $n>1$ is a field if and only if $n$ is prime.
An element $\overline m\in \mathbb Z_n$ with $\overline m\ne \overline 0$ is a zero divisor if and only if $gcd(m,n)\ne 1$
To show that $p|m$ and $p|n$ implies that $\overline m$ is a zero divisor, choose $d:=\frac{n}{p}$ , then you have $\overline m\cdot \overline d=\overline {md}=\overline 0$
The (additive) order is an unusual way to verify whether an element is a zero-divisor. 
But if a non-zero element $m$ has order smaller than $n$, then there is a number $d$ with $0<d<n$ and $\overline m\overline d=\overline md =\overline 0$ , so $m$ is a zero-divisor.
