Rings in which each element is a sum of $n$ commuting idempotents

Question

Let $n$ be a nonnegative integer. Let $R$ be a nonunital ring such that every element of $R$ is a sum of $n$ pairwise commuting idempotents. (As usual, the class of nonunital rings includes the class of unital rings.)

In Theorem 1 of https://math.stackexchange.com/a/1258821 (scroll down to Section 2 for the theorem), I have shown that $\left( n+1\right) !x=0$ for all $x\in R$. What else can be said about $R$ ?

Question 1. Is $R$ commutative?

This question has a positive answer for $n \leq 1$ (this is Stone's famous theorem that every Boolean ring is commutative).

I would be quite happy with an answer that assumes $R$ to be unital. In fact, I think I have an argument showing that if the answer is positive for all unital rings $R$, then it is also positive for all nonunital rings $R$.

EDIT: Will Sawin seems to have discussed this in https://mathoverflow.net/a/142506 , though somewhat too telegraphically to fully understand (at least for me).

Remark. It is tempting to conjecture that $x^{n+1} = x$ for all $x \in R$. And this conjecture indeed holds for $n \leq 1$ (obviously) and for $n = 2$ (see Section 1 of https://math.stackexchange.com/a/1258821 for the proof). But it fails for $n = 3$. Indeed, the ring $\mathbb{Z} / 4 \mathbb{Z}$ has the property that every of its elements is a sum of $3$ pairwise commuting idempotents, but its element $x = 2$ does not satisfy $x^m = x$ for any $m > 1$.

Question 2. Are there integers $a$ and $b$ (depending on $n$ but not on $R$ and $x$) with $a > b > 0$ such that every $x \in R$ is guaranteed to satisfy $x^a = x^b$ ?

Answer 1

The answer to Question 2 is yes (but for uninteresting reasons).

Fix $n$. Let $N=(n+1)!$ and consider the ring $T=\mathbb Z_N[t]$. Let $K$ be the ideal of $T$ generated by $p(t)=t(t-1)\ldots(t-n)$. Since $p(t)$ is monic, $T/K$ is finite. Therefore there exist integers $a>b>0$ with $\bar t^a=\bar t^b$, where $\bar t=t+K\in T/K$.

Now suppose $R$ satisfies the given condition, and consider any $x\in R$. By your linked answer we have $NR=0$ and $p(x)=0$. Let $R^1=\mathbb Z_N\oplus R$ with multiplication $(n,r)(m,s)=(nm,ns+mr+rs)$. This is a unital ring and $NR^1=0$, so we have obtain a homomorphism $T/K\to R^1$ with $\bar t\mapsto x$. Thus $x^a=x^b$.