Given M, can we find $2$ primes $a,b$ so that for all naturals $x,y$, $|a^x-b^y|>M$? For example, if $M = 2$, one can show that $3,17$ satisfy the above:
For any naturals $x,y$, $|3^x-17^y|>2$.

Choose $a$ to be a prime larger than $M + 1$, and let $b$ be the first prime congruent to $1 \pmod{2 \cdot a}$, guaranteed to exist by Dirichlet's theorem.  Now $a^x \equiv a \pmod{2 \cdot a}$ and $b^y \equiv 1 \pmod{2 \cdot a}$, so $\vert a^x - b^y \vert \ge a-1 \gt M$.  Using this method, for $M=3$ we have $(a,b)=(5,11)$, $5^x \equiv 5 \pmod{10}$, and $11^y \equiv 1 \pmod{10}$.  In other words $5^x = \dots 5$ and $11^y = \dots 1$, so they differ by at least $4$.
