Solve for $p^a + 1 = 2\cdot q^b$ where $p,q$ are odd primes and $a,b \ge 2$

Question

Now, clearly, $7^2 + 1 = 2\cdot5^2$.

Is this the only solution? How would I prove this? Or if it is not the only solution, what would be the method to find other solutions?

I'm not clear on how to make progress on this question.

Thanks,

-Larry

Answer 1

There is at least two more solutions (found by brute force): $$\begin{align} 41^2+1&=2\cdot29^2\\ 239^2+1&=2\cdot13^4 \end{align} $$