For any natural number $k$, we have $5^{2k} = M_8 + 1$, $5^{2k+1} = M_8 + 5$, $7^{2k} = M_8 + 1$ and $7^{2k+1} = M_8 + 7$, therefore from $8 \mid 7^a - 5^b$ we deduce that $a$ and $b$ are even. Denote $a = 2m$, $b = 2n$, with $m$ and $n$ positive integers. Then $7^{2m} - 5^{2n} = 8p$, i.e. $(7^m - 5^n)(7^m + 5^n) = 8p$, with $p$ a prime.
If $p = 2$, then $(7^m - 5^n)(7^m + 5^n) \neq 16$.
If $p \ge 3$, from $7^m - 5^n < 7^m + 5^n$, as $7^m - 5^n$ and $7^m + 5^n$ are even, we have the following situations:
$$(1^\circ) \begin{cases} 7^m - 5^n = 4 \\ 7^m + 5^n = 2p \end{cases}, \qquad (2^\circ) \begin{cases} 7^m - 5^n = 2 \\ 7^m + 5^n = 4p \end{cases}.$$
Case (1°): From $7^m = M_3 + 1$ and $5^n = M_3 \pm 1$, it follows that $7^m - 5^n = M_3$ if $n$ is even, and $7^m - 5^n = M_3 + 2$ if $n$ is odd. Since $4 = M_3 + 1$, there are no solutions in this case.
Case (2°): By subtracting the equations, we find that $2p = 7^m - 1$. Since $3 \mid 7^m - 1$, it follows that $3 \mid 2p$, thus $p = 3$. We deduce that $m = n = 1$, $p = 3$, therefore the solution is $(a, b) = (2, 2)$.