Given 2 primitive pythagorean triples with parameters as per: https://en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple
First triple has parameters $(m,n)$. Second has parameters $(r,s)$. Note that $m>n$ and $r>s$. We define $g$ and $h$ according to these equations:
$$g^2 = (mn(r^2-s^2))^2 + (mn(r^2-s^2)-rs(m^2-n^2)+2mnrs)^2$$ $$h^2 = (rs(m^2-n^2))^2 + (mn(r^2-s^2)-rs(m^2-n^2)+2mnrs)^2$$
I'm trying to show that $g$ and $h$ can't both be integers unless $m=r$ and $n=s$.
If you're interested where this comes from, I was working on the rational distance problem: http://unsolvedproblems.org/index_files/RationalDistance.htm
And it got down to proving the above problem. Appreciate any tips on solving the above problem.
This is NOT a complete answer, since I believe that proving this would be extraordinarily difficult. Finding a possible counterexample is also highly non-trivial.
We can simplify the problem by defining $e=m/n$ and $f=r/s$, so we deal with rational variables. Both equations are of the form "quartic in $e$ equals a square". The first equation becomes
\begin{equation*} \Box=f^2e^4-2f(f^2+2f-1)e^3+2(f^4+2f^3-f^2-2f+1)e^2+2f(f^2+2f-1)e+f^2 \end{equation*}
Since $e=0$ gives a solution, this quartic is birationally equivalent to an elliptic curve. Using standard methods, we find the curve
\begin{equation*} V^2=U(U-2f^2(f+1)^2)(U-2(f^2+1)(f^2+2f-1)) \end{equation*} with the reverse transformation \begin{equation*} e=\frac{2f^6+8f^5+8f^4-f^2(U+2)-2f\,U+U-V}{2f(f^4+4f^3+4f^2-U-1)} \end{equation*}
The curves, for fixed $f$, have rank at least $1$, with a point of infinite order at $(\,4f^2(f^2+1), 4f^2(f-1)^2(f^2+1)\,)$ assuming $|f| \ne 0,1$.
Thus one possible method to find a solution is to select $f$, find the generators of the elliptic curve, get points on the curve and find $e$. Then test whether $e$ and $f$ satisfy the second equation.
The second equation also gives a quartic in $e$, namely \begin{equation*} \Box=2f^2e^4-2f(f^2+2f-1)e^3+(f^4+4f^3-2f^2-4f+1)e^2+2f(f^2+2f-1)e+2f^2 \end{equation*} which can also be shown to be equivalent to a different elliptic curve, since $e=1$ gives a solution. This elliptic curve also has rank at least $1$ for fixed $f$.
I hope this small contribution helps in some way.