Berkeley Math Circle

Solution:
The only solution is $(1, 1)$.

It is clear that if $b = 1$ then $a = 1$, and that $(1, 1)$ is a solution. Consequently, assume $b > 1$. Then $6^{b}$ is divisible by $4$. On the other hand, since $5^{a} \equiv 1^{a} = 1 \pmod{4}$ for all $a$, the left side is $2 \pmod{4}$. Thus there are no solutions for $b > 1$.