Claim Both $m,n$ can not be even.
Proof $x+y+z=0$ , $\implies x=-(y+z)$ .
Since $\frac{S_{m+n}}{m+n} = \frac{S_m S_n}{mn}$ ,
by equating cofficient of $y^{m+n}$ on LHS and RHS ,get
$\frac{2}{m+n}=\frac{4}{mn}$ .
$\implies \frac{m}{2} + \frac {n}{2} = \frac{m\cdot n}{2\cdot2}$ .
So we have, \frac{m}{2} \biggm{|} \frac{n}{2} and \frac{n}{2} \biggm{|} \frac{m}{2} .
$\implies m=n=4$ .
So we have $S_8=2(S_4)^2$ .
Now since it will true for all real $x,y,z,x+y+z=0$ .
So choose $x=1,y=-1,z=0$ .
$S_8=2$ and $S_4=2$ so $S_8 \neq 2 S_4^2$ .
This is contradiction. So, at least one of $m,n$ must be odd. WLOG assume $n$ is odd and m is even. The coefficient of $y^{m+n-1}$ in $\frac{S_{m+n}}{m+n}$ is $\frac{\binom{m+n}{1} }{m+n} =1$
The coefficient of $y^{m+n-1}$ in $\frac{S_m\cdot S_n}{m\cdot n}$ is $\frac{2}{m}$ .
Therefore, $\boxed{m=2}$ .
Now choose $x=y=\frac1,z=(-2)$ . (sic)
Since $\frac{S_{n+2}}{2+n}=\frac{S_2}{2}\frac{S_n}{n}$ holds for all real $x,y,z$ such that $x+y+z=0$ .
We have $\frac{2^{n+2}-2}{n+2} = 3\cdot\frac{2^n-2}{n}$ . Therefore,
\begin{equation*}
\label{eq:l2}
\frac{2^{n+1}-1}{n+2} =3\cdot\frac{2^{n-1}-1}{n}\ldots
\tag{**}
\end{equation*}
Clearly $(\ref{eq:l2})$ holds for $n\in\{5,3\}$ .
And one can say that for $n\ge 6$ , $\text{RHS of (\ref{eq:l2})}<\text{LHS of (\ref{eq:l2})}$ .
So our answer is $(m,n)=(5,2),(2,5),(3,2),(2,3)$ .
-ftheftics (edited by integralarefun)