diophantine equation $(1+\ldots+n)^k=1^l+\ldots+n^l$

Question

Here's my equation I wrote and can't solve for over 3 months: $$(1+\ldots+n)^k=1^l+\ldots+n^l$$ in positive integers of course.

For any $n$ there are two obvious solutions: $k=l=1$ and $k=2,\ l=3$. Are there any others?

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I tried to use some congruences, e.g. LHS is divisible by $n$ for odd $n$, but without any sensible result.

Answer 1

It is easy to see that $(1+2+3,\cdots+n)^1=(1^1+2^1+3^1,\cdots+n^1)$ and, for the sums of integers $I(n)$ and cubes $C(n)$

$$I(n)=(1+2+3,\cdots+n)=\frac{n(n+1)}{2}\implies (I(n))^2=\frac{n^2(n+1)^2}{2^2}$$ $$C(n)=(1^3+2^3+3^3,\cdots+n^3)=\frac{n^2(n+1)^2}{2^2}$$ $$\therefore (1+2+3,\cdots+n)^2=(1^3+2^3+3^3,\cdots+n^3)$$

$$\text{In all cases }\quad (1+2+3,\cdots+n)^k=\frac{n^k(n+1)^k}{2^k}$$

The sums of the first $10$ powers are shown here and I have worked out formulas up to $\sum_{x=1}^n x^{31}$.

None of these are even divisible by any power of $I(n)$ except for $l-3$ so I do not believe that are any other solutions.