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
Since $e=0$ gives a solution, this quartic is birationally equivalent to an elliptic curve. Using standard methods, we find the curve
with the reverse transformation
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
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.