Non-expert disclaimer here. 

Applying some heuristics, you get that there are $O(\log N)$ primes of the form $a^m-b^n$ for
fixed $a$ and $b$.

To see this, let's count the values taken by $a^m-b^n$ in the range $[0,N]$. I claim there are $O(\log^2N)$ such terms. If $m$ and $n$ are in the range $[0,2\log N]$, this accounts for $O(\log^2N)$ pairs, a good fraction of which are in the given range.

If $m > 2\log N$, then there is at most one $n$ such that $a^m-b^n$ is in the range. To complete the bound, we need to give an upper bound on the $m$'s for which there can exist $n$'s such that $a^m-b^n$ is in the range.

Consider $m > 2\log N$ and assume that $|a^m-b^n| < N$ so that $n \gtrsim \log N$ also. Notice that $|a^m-b^n|=a^m|1-b^n/a^m|\approx a^m\log(b^n/a^m)=a^m(n\log b-m\log a)$. By Baker et al's results on logarithmic forms, $|n\log b-m\log a|>m^{-k}$ for a constant $k$ that depends on $a$ and $b$, so that $|a^m-b^n|\gtrsim a^m/m^k$. In particular, if $a^m > N(\log N)^k$, you can't be in range. This shows that there are at most $\log^2N$ terms of the form $a^m-b^n$ in range. Of those, how many are prime?

The heuristic would be that the "density" of difference powers is $\log N/N$ (the derivative of $\log^2N$). The density of primes is $1/\log N$ by the PNT. So since they are `obviously' independent, the density of primes of this form is $1/N$. So... there should be infinitely many according to this heuristic, but there should be $O(\log N)$ such primes up to $N$.