This is way too addictive, so I'm going to try to quit, and I'll just leave my thoughts here in case they're useful for other addicts. This is based on the ideas in the previous non-answer that I posted last year, so I'm just editing that answer.

Denote the number of ideals of $R$ by $n(R)$, and define the Alexeq Quotient $q(R)$ to be $\frac{n(R)}{\vert R\vert}$, so the question asks about rings $R$ with $q(R)=1$.

$q(R)$ is multiplicative on direct products (i.e., $q(R\times S)=q(R)q(S)$), so one approach to finding rings with $q(R)=1$ is to look for examples with fairly simple fractions as $q(R)$ and take products.

For example, $q(\mathbb{F}_p)=\frac{2}{p}$, and $q(\mathbb{F}_4)=\frac{1}{2}$, so if we could find a ring with $q(R)=\frac{m}{2^k}$, where $m$ has at least $k$ prime factors (counted with multiplicity) then by taking a suitable direct product with fields we could achieve $q(R)=1$.

It's not hard to find $\mathbb{F}_2$-algebras with $q(R)>1$. For example, take $R$ to be the $d$-dimensional algebra $\mathbb{F}_2\oplus V$ where $V$ is a $(d-1)$-dimensional square-zero ideal for $d>3$.

However, $n(R)$ is bounded by the number of subspaces of $R$, which behaves roughly as a constant times $2^{d^2/4}$ for large values of $d=\dim(R)$. But the number of prime factors of a "typical" large number $n$ is roughly $\log\log n$, giving roughly $\log d$, so unless we can carefully design $R$ so that $n(R)$ has many prime factors, we probably need to be lucky to find an $\mathbb{F}_2$-algebra where $n(R)$ has as many prime factors as $\vert R\vert$.

It's possible to find rings where $n(R)$ has more prime factors than $\vert R\vert$. For example, $R=\mathbb{F}_q[x,y]/(x^2,y^2)$ has $n(R)=q+5$ and so $q(R)=\frac{q+5}{q^4}$. So if $q$ is a reasonably large prime of the form $2^k-5$ (e.g., $q=59$) then $n(R)$ has $2^k$ prime factors, but $\vert R\vert$ has only $4$.

Unfortunately, I don't know any way to do this without introducing large prime factors in the denominator of $q(R)$, which are hard to get rid of: it's difficult to design $S$ so that $n(S)$ is divisible by a particular prime, but $n(S)$ doesn't have many fewer prime factors than $\vert S\vert$.

A weakening of the original question, that I don't think I've found an answer to, is:

**Question:** Is there a finite commutative ring $R$ (apart from finite boolean rings), where $n(R)\geq \vert R\vert$ *and* $n(R)$ has at least as many prime factors than $\vert R\vert$?

I found one example that might be useful. If $R=\mathbb{F}_2[x]/(x^5)\oplus \mathbb{F}_2^4$, where $\mathbb{F}_2^4$ is a square zero ideal, then $n(R)=1296=2^43^4$, and so $q(R)=\frac{3^4}{2^5}$. This means that if we could find a $d$-dimensional $\mathbb{F}_3$-algebra $S$ where $n(S)$ has at least $\frac{5d}{4}$ prime factors then we could construct $T$ with $q(T)=1$.