For what pair of integers $(a,b)$ is $3^a + 7^b$ a perfect square. Problem: For what pair of positive integers $(a,b)$ is $3^a + 7^b$ a perfect square.
First obviously $(1,0)$ works since $4$ is a perfect square, $(0,0)$ does not work, and $(0,1)$ does not work, so we can exclude cases where $a$ or $b$ are zero for the remainder of this post.
I have a few observations made but not much room for a full solution. 
First, since powers of an odd number are odd, and the sum of two odd numbers is even, so the base of the square must be an even number.
Secondly, the last digit of the powers of $3$ are $\{1,3,7,9 \}$ , whereas the last digits of the powers of $7$ are $\{7,9,3,1 \}$.
From here I am not sure how to proceed and any hints much appreciated. I'm not sure if there a finite amount of pairs or not either.

Here is a solution using less machinery than in Will Jagy's answer.
As  Daniel Robert-Nicoud noted in comments, the fact that $3^a+7^b\equiv(-1)^a+(-1)^b$ mod $4$ implies $a$ and $b$ must have opposite parity in order for the (even) sum $3^a+7^b$ to be a perfect square. So we seek to show that $(a,b)=(1,0)$ is the only solution in the (odd,even) case and $(a,b)=(2,1)$ is the only solution in the (even,odd) case.
If $a$ is odd and $b=2m$ is even, then we can rewrite $3^a+7^b=n^2$ as $3^a=(n+7^m)(n-7^m)$.  This implies $n+7^m$ and $n-7^m$ are each powers of $3$, say $3^c$ and $3^d$ with $c+d=a$.  But then $2\cdot7^m=3^c-3^d$, so we must have $d=0$ and $c=a$, since $3\not\mid2\cdot7^m$.  One solution to $2\cdot7^m=3^a-1$ is $(a,m)=(1,0)$.  It remains to show there are no solutions with $m\gt0$.  For this it suffices to note that $3^a\equiv1$ mod $7$ if and only if $6\mid a$, which contradicts the assumption that $a$ is odd.
In the other parity case, if $a=2m$ is even and $b$ is odd, then, along similar lines, we must have $2\cdot3^m=7^b-1$, which has $(m,b)=(1,1)$ as one solution.  After checking that there is no solution with $m=0$, it remains to show there are no solutions with $m\gt1$.  If there were, then we would have $7^b\equiv1$ mod $9$, which would imply $b\equiv3$ mod $6$ (since $6$ is the order of the multiplicative group of units mod $9$ and $7^3\equiv(-2)^3=-8\equiv1$ mod $9$). Writing $b=3r$ (with $r$ odd, but that doesn't matter), we note that $(7^3-1)\mid(7^{3r}-1)$, so that $7^3-1=2\cdot3^2\cdot19$ divides $2\cdot3^m$, an obvious contradiction.
