In an answer to my own question, I showed that if $p$ is an odd prime and $p=a^m \pm b^n$ with positive integers $a,b$ relatively prime to $p$, then $p$ does not simultaneously divide the Fermat quotients $\frac{a^{p-1}-1}{p}$ and $\frac{b^{p-1}-1}{p}$.

Once upon a time, this result would have been of some interest because it implies the first case of Fermat's Last Theorem (FLT) for certain prime exponents $p$. More precisely, Wieferich, Mirimanoff, and others showed that if the first case of FLT fails for the prime $p$, then $\frac{a^{p-1}-1}{p}$ will be divisible by $p$ for every prime number less than or equal to 89.

In particular, the above two paragraphs show that the first case of FLT holds if $p = a^m \pm b^n$ for some primes $a, b \leq 89$.

Although FLT was proved by independent methods, results like the one in my other question still have some interest. For instance, as far as I know, it is still unknown whether there are nontrivial solutions $x,y,z$ in the cyclotomic field $\mathbb{Q} (\zeta_p)$ (whether they are all relatively prime to $p$ or not) that satsify

$$ x^p + y^p = z^p$$

In ``On the first case of the fermat theorem for cyclotomic fields", Kolyvagin proves that there are no such solutions relatively prime to $p$ if each of the numbers $\frac{a^{p-1}-1}{p}$ with $a$ prime and $\leq 89$ is relatively prime to $p$. In particular, this is true when $p$ has the form $p=a^m \pm b^n$ with $a, b$ primes $\leq 89$.

I would presume that the set of such primes is quite scarce (although the possibility $p = a^m-b^n$ throws me a bit), so my question is:

Are there known heuristics or results for the set of primes $p=a^m \pm b^n$ with $a,b$ primes less than $89$ (one of them, of course, being 2)?