<s>[INST] Find all nonnegative integers 
Determine all nonnegative integers $x$ and $y$ so that $$3^x + 7^y$$
  is a perfect
  square and $y$ is even.

Without trial-and-error of course. 
$$3^x + 7^y = a^2$$ For some integer $a$. 
$$\implies a = \sqrt{3^x + 7^y}$$
But is there really a way?
 [/INST] Writing $y = 2z$ we have $3^x = (a + 7^z)(a - 7^z )$.  Now $a + 7^x$ and $a - 7^x$ must be two powers of $3$, say $a + 7^x = 3^d$, $a - 7^x = 3^e$.
But $3^d - 3^e = 2 \times 7^x \equiv 2 \mod 3$ so we must have $e = 0$.  Then
$a = 1 + 7^x$, and  $3^d = a + 7^x = 1 + 2 \times 7^x$.  One solution is $x = 0$, $d = 1$, corresponding to $3^1 + 7^0 = 2^2$.  Otherwise, note that $3^d \equiv
1 \mod 7$ iff $d$ is divisible by $6$, but if $d$ is divisible by $6$ then $3^d \equiv 1 \mod 4$ as well, so $(3^d - 1)/2$ is divisible by $2$ and can't be a power of $7$.  So there are no other solutions.
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