Below claim is consequence to Erdős–Moser equation
Define $S_m(n)=1^m+2^m+...+n^m$
Is there any solution for $S_m(x)=y^m$ other than $x=1,y=1$ where $x,y,m\in \mathbb{Z}_+$ and $m>2$
I already asked, solution of $m=2$, check here there are only two solution.
generalized version of the above problem, check here
Thanks for your answer and references in advance.*
On
$m=4$
$$30y^4=x(x+1)(2x+1)(3x^2+3x+1).$$ The factors $x$ and $x+1$ are coprime to each other and to the other factors. Furthermore, one of $x$ and $x+1$ is even. Therefore $\{x,x+1\}=\{2au^4,bv^4\}$, where $ab \in \{1,3,5,15\}$
Each of these relatively few possibilities gives a Thue equation. For example, consider $$2u^4-3v^4=\pm1.$$ Checking using PARI/GP gives only the trivial solutions $x=\pm 1,y=\pm 1$. The other possibilities can be checked in the same way and (this was a quick check so needs proper verification) the computer found there to be no non-trivial solutions. (And this was without even considering the other factors!)
The general case
We have seen that just using the factors $x$ and $x+1$ gives us a way into these equations. You may be able to use Faulhaber's formula to obtain some general results about these factors.
I hope this is of some use/help.
$m=2$
There are proofs without using elliptic curves mentioned in the Wikipedia article for which you already have a reference. I can remember proving this result some years ago - most of the proof involved a straightforward use of Fermat's method of infinite descent except for one awkward case which, if my memory is correct, involved a clever use of quadratic reciprocity which I had seen in someone else's paper.
$m=3$
(Please ignore if you have already dealt with this case).
We have $x^2(x+1)^2=4y^3$. So $y$ is a square, $u^2$ say, and then $x(x+1)=2u^3$. Let $r=2x+1$ and $s=2u$, then $$r^2=s^3+1.$$
This is an elliptic curve with integer points only if $s\in \{-1,0,2\}$. Then $y$ is $0$ or $1$ and the only solution is $x=y=1$.