When is $1^n+2^n+3^n+4^n+\cdots+n^n$ a square number? I've been thinking about such an interesting question these last few days. I wonder if anyone has studied it.

Question:  Find all postive integers $n$ such that
  $$S(n)=1^n+2^n+3^n+4^n+\cdots+n^n$$ is a square number?

Here $n=1$ is clear and if $n=3$ then $S(3)=36=6^2$. Now I can't find any other $n\le 20$, so I conjecture these $n$ are the only two values.
 A: An easy obstruction (mod 8). We can exclude all odd multiples of $4$, because for any $k\in\mathbb{N}$, $S(8k+4)=2\mod 4.$  (Reason: if $j$ is even, $j^{8k+4}=0\mod 4$, and  if $j$ is odd, $j^{8k+4}=1\mod 4$, and there are $4k+2$ odd numbers from $1$ to $8k+4$).
A: This is not an answer, but I don't have enough reputation to comment. For large $n$, $S(n)\approx n^n.$ The density of square integers at $n^n$ is approximately $n^{-n/2}$. So one might expect that the number of $S(n)$ which are square is about $\sum_{\mathbb N} n^{-n/2} \approx 1.7788$. 
A: A special case,
Thm 1:    if prime $\ p\equiv 3\ $ mod $4,\ $ then
$\ 1^{p-1}+\ldots (p-1)^{p-1}\ $ is not a square.

A: Not a real answer either, but this link may be relevant : Faulhaber's formula
In particular one could try to establish, following Pietro Majer and Gerhard Paseman, that a solution of the diophantine equation  $ S(n)=m^{2} =P_{n}(a)$ only occurs for odd values of  $ n $, where  $ a=n(n+1)/2 $ and  $ P_{n} $ is the  $ n $-th Faulhaber polynomial.
