We will prove the inequality $5^{n+1} \ge n^6$ holds for all positive integers $n$, except $n = 3$ and $n = 4$.

Indeed, this is readily verified for $n = 1$, $n = 2$ and $n = 5$. When the thesis is true for $n \ge 5$, then $5^{n+2} = 5 \cdot 5^{n+1} \ge 5n^6$. It is then sufficient to show that $5n^6 \ge (n+1)^6$, rewritten as $5 \ge \left(1 + \frac{1}{n}\right)^6$. Since $\left(1 + \frac{1}{n}\right)^6 < \left(1 + \frac{1}{4}\right)^6 = 5 \cdot \frac{5^5}{4^6} = 5 \cdot \frac{3125}{4096} < 5$, everything is thus proven, by simple induction; moreover, the inequality becomes strict for $n > 5$.

One thus gets $5^{a+1} \ge a^6$ and $5^{b+1} \ge b^6$, for all $a, b$ different from $3$ and $4$. Multiplying, first the two inequalities given in the problem statement, then those two just obtained in the above, we get $a^6 b^6 \ge 5^{b+1} 5^{a+1} \ge b^6 a^6$. But equality only holds here if $a = b = 5$, which clearly checks. Otherwise, it needs $a = 4$ or $a = 3$ (or, symmetrically, $b = 4$ or $b = 3$). If $a = 4$, then the inequalities given in the problem statement become $4^6 > 5^{b+1}$ and $b^6 > 5^5$, leading to $b \le 4$ and $b \ge 4$ respectively, hence $b = 4$. If $a = 3$, then $3^6 > 5^{b+1}$ and $b^6 > 5^4$, leading to $b \le 3$ and $b \ge 3$ respectively, hence $b = 3$.

Two more pairs of solutions $a = b = 4$ and $a = b = 3$ have thus been found.