Idk if this has anything to do with this problem but it seems kind of related so I'll just post it here. https://www.overleaf.com/project/5c64c1d24aaafd4a547f0e03

A solution with a bit of calculus. The answer is all $(m,n)=(2k+1, 2t(2k+1))$ for $k,t \in \mathbb {Z}$ and $t >0$. Now we prove that these are the only $m,n$ which satisfy. Case 1) Both $m,n$ are even. It is easy to see that, $(x,y)=(1,1)$ and $(x,y)=(1,-1)$ give the same value $x^n+y^n$ and $x^m+y^m$ but give different values of $xy$. So, such $m,n$ are not possible. Case 2) Both $m,n$ are odd. Take, $(x,y)=(1,-1)$ and $(x,y)=(0,0)$. Case 3) Exactly one of $m,n$ is even. WLOG let $n=2p$ be even. It is pretty easy to see that $p \neq 0$. We claim that $m|p$. Suppose not. We will show the existence of $(x_1,y_1)$ and $(x_2,y_2)$ such that $x_1^{2p} +y_1^{2p} = x_2^{2p} +y^{2p} =a$ and $x_1^m +y_1^m = x_2^m +y_2^m=b$ but $x_1y_1 \neq x_2y_2$. Which will give a contradiction. This is easy, just take $b=2$ and consider the function, $$f(x) =(1-x)^{\frac{2p}{m}} + (1+x)^{\frac{2p}{m}}$$ Unless, $m|p$ this function has a extremum at $x=1$. So, there are $c_1<1<c_2$ in the neighborhood of $1$ so that, $f(c_1)=f(c_2)$.Take $a=f(c_1)$ . Then, It is easy to see that $(x_1,y_1)=((1-c_1)^{\frac{1}{q}},(1+c_1)^{\frac{1}{q}})$ and $(x_2,y_2)=((1-c_2)^{\frac{1}{q}},(1+c_2)^{\frac{1}{q}})$ satisfy. Therefore, $m=2k+1, n=2t(2k+1)$. If $t \leq 0$ we can use the same argument: extremum at $1$. So, $t>0$. To prove that such $(m,n)$ work is almost the same as above. Because, if $x^m+y^m =a \implies x=\left(\frac{a}{2} -t \right)^{\frac{1}{2k+1}}, y=\left(\frac{a}{2} +t \right)^{\frac{1}{2k+1}}$ for some $t \in \mathbb{R}$. But note that the function $g$ (replace $1$ with $\frac{a}{2}$ in $f$) has the property that $g(t_1)=g(t_2) \implies t_1 = \pm t_2$. Which implies a fixed value of $xy$. So, we are done.

The solution above has an error. The function f has an extremum iff n/m is smaller than 1. So the right answer is (2k,2l+1) where k>l.