Fix $m$ and $n$. Writing the $S_k$ in the equation given in terms of $x, y, z, m,$ and $n$ and using $z = -(x + y)$: $$\frac{x^{m+n} + y^{m+n} + (- x - y)^{m+n}}{m+n} = \frac{(x^m + y^m + (- x - y)^m)(x^n + y^n + (- x - y)^n)}{mn}$$ Since $x$ and $y$ are completely arbitrary, we treat this as a polynomial in $x$ and $y$, so we equate coefficients. By the Binomial Expansion, if $m$ and $n$ are both odd, we have a term with $x^{m+n}$ on the LHS, but no $x^{m+n}$ term on the RHS, so at least one of $m$ or $n$ is even. Suppose both are even. Now, we equate the $x^{m+n}$ coefficients on both sides: $$\frac{2}{m+n} = \frac{4}{mn}$$ Plug in $m = 2m'$ and $n = 2n'$. By simplifications and SFFT, $(m' - 1)(n' - 1) = 1$. This means $m' = n' = 2$ and $m = n = 4$. Therefore, $S_8 = 2 \cdot (S_4)^2$. However, this relation must be satisfied no matter what we choose for $x$ and $y$. Choosing $(x, y, z) = (1, -1, 0)$ yields a contradiction. So one of $(m, n)$ is odd and the other is even. WLOG, assume $m$ is even. The $x^{m+n}$ terms cancel on both sides, since $m + n$ is odd, so we look at the coefficient of $xy^{m + n - 1}$. By the Binomial Expansion, the coefficient on the LHS is $-\frac{\binom{m+n}{1}}{m + n} = -1$. For the first factor on the RHS, we have $(2x^m + 2y^m + ....)$ For the second factor, we have $(-\binom{n}{1}xy^{n-1} - [\text{stuff with an x power greater than 1}])$. So the coefficient on the RHS is $\frac{-2}{m}$. So $-1 = \frac{-2}{m}$ or $m = 2$. Now we have $\frac{S_{n+2}}{n + 2} = \frac{S_2 \cdot S_n}{2n}$. Choose $(x, y, z) = (2, -1, -1)$. Plugging the stuff into the formula (remember $n$ is odd) gives us: $\frac{2^{n+2} - 2}{n + 2} = 6 \cdot \frac{2^n - 2}{2n}$ Obviously, the LHS will grow faster than the RHS. In particular, when $n \geq 6$, the LHS will overtake the RHS. So there are no other candidates for solutions besides the ones given.