Once you have this, at least one is $0\mod 5$ and at least one is $0\mod 17$
In $\mathbb{Z}_5$ we have $\sqrt{-1}=\pm 2$. In $\mathbb{Z}_{17}$ we have $\sqrt{-1}=\pm 4$.
If both factors are $0\mod 5$ then the first factor gives $n\in\{3,4\}\mod 5$ and the second factor gives $n\in\{1,2\}\mod 5$. Contradiction.
If both factors are $0\mod 17$ then the first factor gives $n\in\{5,14\}\mod 17$ and the second factor gives $n\in\{3,12\}\mod 17$. Contradiction.
So we have two cases:
1. $n^2-2n+2=1$ and $n^2+2n+2=85^m$.
2. $n^2-2n+2=5^m$ and $n^2+2n+2=17^m$.
Case 1 gives $n=1$ but $1^2+2\cdot 1+2=5\neq 85^m$. So this case won't work.
Case 2:
$\frac{n^2+2n+2}{n^2-2n+2}\geq\frac{17}{5}\implies 17n^2-34n+34\leq 5n^2+10n+10\implies 12n^2-44n+24\leq 0$
$\implies(n-3)(3n-2)\leq 0\implies\frac{2}{3}\leq n\leq 3$
So we check $n\in\{1,2,3\}$ to see that the only solution is $(m,n)=(1,3)$.