Find $a,b \in \mathbb{Z}^+$ such that : $\frac{a^{2}-2}{ab+2}\in \mathbb{Z}$ 
$1$. Find $a;b\in \mathbb{Z}^+$ such that : $\frac{a^{2}-2}{ab+2}\in \mathbb{Z}$
$2$. Find $m;n>1$ such that : $2^m+3^n=k^2$ $(k\in \mathbb{Z})$

Problem 1. I thought :
$\frac{a^{2}-2}{ab+2}\in \mathbb{Z}\Rightarrow a^{2}-2\vdots ab+2\Rightarrow a^{2}+ab\vdots ab+2\Rightarrow ab(a+b)\vdots ab+2\Rightarrow 2(a+b)\vdots ab+2$
But i don't know how to do next!!

1
Case 1 : $\frac{a^2-2}{ab+2} =0$ no solution
Case 2 : $\frac{a^2-2}{ab+2} >0 $ and $a < b$ : So $$
\frac{a^2-2}{ab+2} < 1$$
no solution.
Case 3 : $\frac{a^2-2}{ab+2} >0 $ and $a \geq b\ (a>1,\ a>b)$ : So $$
\frac{a^2-2}{ab+2} < \frac{a^2}{ab} = \frac{a}{b},\ ba^2-2b < a^2b +2a$$
So $$ ab\leq ab(a-b)< 2(a+b),\ (a-2)(b-2)< 4$$
Hence if $b \geq 3$, then $a\leq 5$. In this case by computation we have $(a,b)=(4,3)$ (There was a solution-finding of questioner)
Case 4 : $\frac{a^2-2}{ab+2} <0 $ : no solution.
