Let $n > 2$ be an integer. I've got a pair of equations
$$x^2 = (n^2 - 2)y^2 \pm 1,$$
one of which has a solution when $y=n$ (with +1); I'm trying to say that, when $1 \leq y < n$, neither equation has a solution.
André's link shows that the $-1$ case never has solutions, so we can focus on the $+1$ case. We can see that $(n^2-2)y^2+1 = n^2y^2 - 2y^2+1 $ is less than $n^2y^2$ when $y\geq 1$ as we assumed. We can also check that $$(n^2-2)y^2+1 > (ny-1)^2 $$ by writing a sequence of equivalent inequalities until we see one to be true. Expanding gives:$$ n^2y^2 - 2y^2 + 1 > n^2y^2 - 2ny + 1 .$$ Subtracting common terms leaves $$-2y^2 > - 2ny $$ and that is equivalent to $y<n$ which is assumed. Thus if $1\leq y<n$ then $(n^2-2)y^2+1$ lies between two consecutive squares, and thus can not be a perfect square itself.