By just considering $S_{n+1}-S_n=\frac{2^{n+1}}{(n+1)^2}$ we see immediately that this is equivalent to $\frac{S_n}{2^n}$ being a rational function. Denote $\frac{S_n}{2^n}=r(n)$ for some rational function $r$. Now \[\frac{2}{(n+1)^2}=2r(n+1)-r(n).\]This is true for all positive integers $n$ and hence must be true as an identity of rational functions identically. In particular, note that $r(1)=1$ and $r(0)=0$ and that $r$ must have a pole at $x=-1$. But then $r$ must have a pole at $x=-2$, then at $x=-3$ etc. so that $r$ must have infinitely many poles which is clearly a contradiction.