Other variation of Nicomachus's Theorem?

We all know that $ 1^3+2^3+3^3 + \ldots + n^3 = (1+2+3+\ldots + n)^2 $.

Denote $\displaystyle S_m = \sum_{j=1}^n j^m $, then we can set $ S_3 = S_1 ^2 $ for all positive integers $ n $.

Question: Is there any other solution for the equation $ S_a= (S_b)^c $ for distinct positive integers $a, b $ and rational number $c$?

(Yes, this means that I'm searching for an algebraic identity that satisfies all positive integers $n$.

My feeble attempt: By method of differences, we can see that the leading coefficient of the polynomial $S_m $ is $ \frac1{m+1} $, with that, we get $ \frac1{a+1} = \left( \frac1{b+1} \right)^c $ or $(a+1)=(b+1)^c $. And I'm stuck.

I tried with Faulhaber's formula but nothing seem to work.

I'm pretty sure there's no other solution but there's no solid proof, and even if there is, $a$ and $ b$ must share the same parity.

Any help would be greatly appreciated! Thank you.

This is a partial solution.

Theorem : For positive integers $a,b,c$ where $a\not=b$, if $S_a=(S_b)^c$ holds for all positive integers $n$, then $(a,b,c)=(3,1,2)$.

Proof : Note that one has $$\frac 1n\sum_{k=1}^n\left(\frac kn\right)^a=\frac{n^{(b+1)c}}{n^{a+1}}\left(\frac 1n\sum_{k=1}^n\left(\frac kn\right)^b\right)^c.$$ Considering the limit $n\to\infty$ on the both sides gives us$$\frac{1}{a+1}=\left(\frac{1}{b+1}\right)^c\ \ \ \ \text{and}\ \ \ \ (b+1)c=a+1\tag1$$ Hence, one has $$c=(b+1)^{c-1}.$$ If $c=1$, then $a=b$, which we exclude.

If $c=2$, then $b=1,a=3$.

If $c\ge 3$, then $$\ln(b+1)=\frac{\ln c}{c-1}$$ has no solution because $f(x)=\frac{\ln x}{x-1}$ is decreasing for $x\ge 3$ and $f(3)=\frac{\ln 3}{2}\lt\ln 2$.

Hence, $(a,b,c)=(3,1,2)$ is the only solution. Q.E.D.