51st Ukrainian National Mathematical Olympiad, 3rd Round

Answer: $(2,3)$, $(3,2)$, $(2,2)$.

Clearly, either $a$ or $b$ is even. WLOG, $a=2$. It is easy to see that $b=2$ and $b=3$ satisfy the condition. Suppose that $b>3$. We have: $2^b=b^2+1$, $b>3$. It is easy to see that the last expression is divisible by $3$ and cannot be prime.