Every nonzero element in a finite ring is either a unit or a zero divisor

In a finite commutative ring with unity, every element is either a unit or a zero-divisor. Indeed, let $a\in R$ and consider the map on $R$ given by $x \mapsto ax$. If this map is injective then it has to be surjective, because $R$ is finite. Hence, $1=ax$ for some $x\in R$ and $a$ is a unit. If the map is not injective then there are $u,v\in R$, with $u\ne v$, such that $au=av$. But then $a(u-v)=0$ and $u-v\ne0$ and so $a$ is a zero divisor.

For the noncommutative case, see this answer.

Your question is incomplete: you say you want to prove that every nonzero element of $R$ is "either a zero-divisor?" If one inserts a unit or before zero-divisor then you get a true statement, so I'll assume for now that's what you meant.

First, following a comment by Gerry Myerson on a recent related answer, let me divulge that for me zero is a zero-divisor. I claim that this is just a convention that you should be able to translate back from if you see fit.

Next, note that if you have a family $\{R_i\}_{i \in I}$ of rings in which every element is either a unit or a zero-divisor, the same holds in the Cartesian product $R = \prod_{i \in I} R_i$.

In your case you can use the structure theorem for Artinian rings: $R$ is a finite product of local Artinian rings -- to reduce to the case in which $R$ is local Artinian. Then the maximal ideal is nilpotent, so every nonunit is nilpotent and in particular a zero divisor.

Hint $\,\ \overbrace{|R|<\infty\ \Rightarrow\ r^j=r^k}^{\rm\large pigeonhole},\: j>k\ $ $\Rightarrow\ (r^{j-k}-1)\,\color{#0a0}{r^k}=0\ $ $\overset{\!\large \color{#0a0}{r\ \nmid\ 0}}\Longrightarrow\ \overbrace{r^{j-k}=1}^{\!\!\!\!\textstyle\color{#c00}r\,(r^i)\!=\!1^{\phantom{|^|}}\!\!\!\!\!\!}\, $ $\,\Rightarrow\, \color{#c00}r\, $ is a unit

Remark $\ $ The idea generalizes: if a non-zero-divisor $\,r\,$ is algebraic then it divides the least degree coefficient of any polynomial of which it is a root. When said coefficient is a unit then so too is $\:r.\:$ Hence the result holds more generally for any ring satisfying a polynomial identity whose least degree coefficient is unit, e.g. for Jacobson's famous rings satisfying the identity $\rm\:X^n =\: X\:.$

P. M. Cohn has shown that every commutative ring $R$ can be embedded in a ring $S$ where every element of $S$ is either a zero-divisor or a unit of $R\,$ (he deems this a "rough zero-divisor dual" of fraction / localization extensions)