Find all integers that make this expression rational I came up with this difficult problem a while ago while solving another relatively easy problem. 
Find all integers m and n, such that $m^2 + n^2$ is a square, and such that  $\sqrt{\frac{2m^2+2}{n^2+1}}$ is rational. 
I've already tried the pythagorean substitution numerous times to no avail and to make things worse it isn't even homogenous. Come to think of it I must have made a mistake in reducing my problem to this. 

Too long for a comment. As requested by OP as a sidenote, if we relax the requirements on $m,n$ and allow rationals, then the system,
$$m^2+n^2 = x^2,\quad \frac{2(m^2+1)}{n^2+1}=y^2\tag1$$
does have an infinite number of rational solutions. Other than the obvious case $n=1$, we have two quadratic polynomials in $m$ to be made squares,
$$f(m) = x^2,\quad\quad g(m) = y^2$$
hence an intersection of two quadric surfaces. A simple rational solution to $(1)$ is then,
$$m = \frac{(a-c)c}{(a-2c)b},\quad n=\frac{a}{2b}$$
where,
$$a^2+2b^2 = c^2\tag2$$
and $(2)$ is easily solved as $a,b,c = u^2-2v^2,\, 2uv,\, u^2+2v^2$ yielding,
$$x^2=\frac{(u^4+4u^2v^2+20v^4)^2}{(4uv)^2(u^2+6v^2)^2},\quad y^2 = \frac{2(u^2+4v^2)\,(4v)^2}{(u^2+6v^2)^2}$$
Thus, it remains to find $u^2+4v^2 = 2z^2$ which can also be solved parametrically. For a particular example, let $u,v,z = 2,\,1,\,4,$ so $m,n = \frac{3}{5},\frac{1}{4},$ and $x,y = \frac{13}{20},\frac{8}{5}$.
P.S. There are no positive integer solutions to $(1)$ with $m,n<500$.
