$x^n = x$ implies commutativity, a universal algebraic proof?

A subdirectly irreducible ring satisfying the equation $x^n=x$ for some $n\geq 2$ is a finite division ring

First let's consider the elementary part of this statement. Observe

Lemma. If $R$ is a subdirectly irreducible ring with no nonzero elements that square to zero, then $R$ has no nontrivial zero divisors.

Reasoning: Assume not. Then $R$ has elements $a\neq 0\neq b$ such that $ab=0$. The set $bRa$ consists of elements that square to zero, so $bRa=\{0\}$. Then $(b)(a)=RbRaR=R\{0\}R=\{0\}=(0)$. If $I$ is the least nonzero ideal of $R$, then $I\subseteq (b)$ and $I\subseteq (a)$, so $I^2\subseteq (b)(a)=(0)$. This shows that all elements of $I$ square to zero, a contradiction. \\\

Here is how the lemma applies. No nonzero element $a\in R$ which squares to zero can satisfy the equation $x^n=x$, so the lemma implies that if $R$ is subdirectly irreducible and satisfies $x^n=x$, then $R$ has no nontrivial zero divisors. Given any nonzero $a\in R$ we have $0=a^n-a = a(a^{n-1}-1)$. Since $a\neq 0$, it must be that $a^{n-1}-1=0$. Hence, any nonzero $a\in R$ satisfies $a^{n-1}=1$, showing that nonzero elements are units. Thus, $R$ is a division ring.

The non-elementary part of the statement

A subdirectly irreducible ring satisfying the equation $x^n=x$ for some $n\geq 2$ is a finite division ring

is that a division $R$ whose nonzero elements satisfy $x^{n-1}=1$ must be finite. The idea for this is to first prove that any maximal subfield of $R$ is finite (which is easy), then to prove that any finite maximal subfield of $R$ has finite index in $R$ (see Lam's A First Course in Noncommutative Rings, Theorem 15.4).

I had the following idea: since $R$ has no nilpotent elements it should be a subdirect product of integral domains satisfying the same equations - however I know this property for commutative rings, and it relies on the well-known fact that $\displaystyle\bigcap\{p, p\in \mathrm{Spec}R\} = \{x, x \text{ is nilpotent}\}$ - and I don't know whether this is true for noncommutative rings.

That much is true for noncommutative rings.

The condition that $x^n = x$ for some $n$ implies $R$ is von Neumann regular, and it is well-known that a reduced VNR ring (which is called a strongly regular ring) is a subdirect product of division rings. Therefore a subdirectly irreducible ring with these properties is a division ring.

The thing working for us here is that in a strongly regular ring, quotients by prime ideals are division rings, so the prime ideals actually satisfy the commutative definition of "prime." As such, their intersection contains all nilpotent elements, and their intersection is zero. So, you can rely on a similar argument.