TITLE: Perfect squares with two variables.
QUESTION [0 upvotes]: Find all positive integers m, n such that $6^m + 2^n + 2$ is a perfect square. 
I've tried keeping a constant value of m and finding out n. 
Eg: 
$m=1, n=0$
$m=1, n=3$

REPLY [0 votes]: As people in the comments mentioned, if $m > 1$ and $n > 1$ than our number divide $2$, but not divide $4$, so it cant be square. So the only cases are:
$1$) $m = 1$
In this case we have equation $2^n + 8$ is a perfect square. But if $n > 3$ than this number divide $8$, but not divide $16$, so it cant be square. So just consider $n = {1,2,3}$
$2$) $n = 1$
In this case we have equation $6^m + 4$ is a perfect square. But $6^m + 4 \equiv$ $3\ or \ 5\ (mod\ 7)$ which is not quadratic residue.
Thats all cases.
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