Determining the structure of a finite ring I'm working on the following problem and would like some guidance

$R$ is a finite ring with $x^5 = x$ for every $x$ in $R$.  Determine the structure of $R$.

My first thoughts were to factor and look at cases whether $R$ has zero divisors.
If $R$ has no zero divisors, then $x^4 = 1$ for nonzero $x$, and so $R$ is a field, namely $\mathbb{Z}/5$
Then you could look at rings of the form $\mathbb{Z}/n$ and focus on the multiplicative group.  But this does not guarantee you'll find all commutative rings and it does not include non-commutative rings. So I think you have to use some ideas with the jacobson radical and semisimplicity.
Source: Fall 1996
 A: Because $xx^3x=x$ for all $x$, the ring is von Neumann regular.
Exercise: $R$ also has no nonzero nilpotent elements.
Exercise: If there are no nonzero nilpotents, that implies all idempotents are central.
Exercise: the idempotents are partially ordered by the relation $e\leq f$ iff $ef=e$.
Exercise: every two idempotents have a common upper bound . Hint:  $e+f-ef$
Exercise: By finiteness, there is a unique maximal idempotent with respect to that order, and it is the identity of the ring (Hint: Let $e$ be a maximal idempotent. For arbitrary $a$, there exists $x$ such that $axa=a$. Notice that $ax$ and $xa$ are idempotents and look first at $ea=eaxa$.)
Exercise: a finite von Neumann regular ring is semisimple (having an identity speeds this along, somewhat.)
Since the ring has no nonzero nilpotents, it is a finite product of division rings. (Using Artin-Wedderburn.)
Since finite division rings are fields, it is a finite product of finite fields. (Wedderburn's little theorem.)
In each such field, since every element is a root of $x^5-x$, each of these fields can have no more than $5$ elements.
You can verify that $F_5, F_3, F_2$ all work, but $F_4$ does not.
So, now you should be able to see completely what such a ring looks like. I leave it to you to put into words.

Alternatively, you can apply a sledgehammer due to Jacobson to say very early on that $R$ is commutative, and find an identity the same way. I tried to avoid that.
