Let $n$ be even, fixed, large.
Sketch of proof that $\{m|m>0, 1^n+\dots+m^n \mbox{ divides } (1+\dots+m)^n \}$ has zero density in $\mathbb{N}$.
Let $f(n)$ be the degree of the largest irreducible factor of $B_{n+1}(x)/(n+1)$ in $\mathbb{Q}[x]$. Let $S$ be the set of primes $p$ for which this factor splits into linear factors mod $p$. $1^n+\dots+m^n$ divides $(1+\dots+m)^n$ only if $m+1$ mod $p$ for each $p\in S$ is in one of the at most $p-(f(n)-2)$ classes mod $p$ for which $B_{n+1}(m+1)/(n+1)$ is not $0$ mod $p$ or both $B_{n+1}(m+1)/(n+1)$ and $(B_2(m+1)-1/6)/2$ are $0$ mod $p$. Since $f(n)>2$ (check literature or prove it), and $S$ has positive density in the primes by Chebotarev's density theorem, the density of these $m$ is zero.