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.

The technique I mentioned is working.   One of the cases is
$$ 1 + 2 \cdot 3^c = 7^d,  $$
where we think that $c=d=1$ gives the largest such answer. Subtract $7$ from both sides,
$$  2 \cdot 3^c - 6 = 7^d - 7.  $$ Let $y+1 = c,$ $x+1 = d,$ for
$$ 6 \cdot 3^y - 6 = 7 \left( 7^x -1 \right), $$
$$ 6 \left( 3^y -1 \right) = 7 \left( 7^x -1 \right).   $$
We are assuming $x,y \geq 1,$ with both sides being nonzero.
$$ 3^y \equiv 1 \pmod 7, $$
$$ y \equiv 0 \pmod 6. $$
Next, $3^y - 1$ is divisible by $3^6 - 1 = 728 = 8 \cdot 7 \cdot 13.$
$$ 7^x \equiv 1 \pmod {13}, $$
$$ x \equiv 0 \pmod {12}. $$
Next, $7^x - 1$ is divisible by $7^{12} - 1 = 13841287200 = 32 \cdot 9 \cdot 25 \cdot 13 \cdot 19 \cdot 43 \cdot 181.$
$$ 3^y \equiv 1 \pmod {43}, $$
$$ y \equiv 0 \pmod {42}. $$
Next, $3^y - 1$ is divisible by $3^{42} - 1 = 109418989131512359208 = 8 \cdot 7^2 \cdot 13 \cdot 43 \cdot 547 \cdot 1093 \cdot 2269 \cdot 368089$
All we care about is that $3^y - 1$ is divisible by $7^2 = 49,$ as this contradicts the assumption of $x \geq 1$ in 
$$ 6 \left( 3^y -1 \right) = 7 \left( 7^x -1 \right).   $$
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