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.

Just to register, from the comment by Daniel, there are just two possibilities, if the result is $x^2,$ either
$$ 1 + 2 \cdot 3^c = 7^d,  $$ or
$$  1 + 2 \cdot 7^e = 3^f. $$
I would guess that an elementary method shown in an answer by 
Gyumin Roh to http://math.stackexchange.com/questions/1551324/exponential-diophantine-equation-7y-2-3x  can be modified for this task. My way of working with this takes a while... 
http://math.stackexchange.com/questions/1941354/elementary-solution-of-exponential-diophantine-equation-2x-3y-7
http://math.stackexchange.com/questions/1941354/elementary-solution-of-exponential-diophantine-equation-2x-3y-7/1942409#1942409 
http://math.stackexchange.com/questions/1946621/finding-solutions-to-the-diophantine-equation-7a-3b100/1946810#1946810 
http://math.stackexchange.com/questions/2100780/is-2m-1-ever-a-power-of-3-for-m-3/2100847#2100847
