If $\left(1^a+2^a+\cdots+n^{a}\right)^b=1^c+2^c+\cdots+n^c$ for some $n$, then $(a,b,c)=(1,2,3)$?

This is a partial solution. I've just been able to get the following theorem:

Theorem : If $(\star)$ for some $n=8k-5,8k-4\ (k\in\mathbb N)$, then $(a,b,c)=(1,2,3)$.

Lemma : For $a\ge 2, n\ge 2,$ $$\left(1^a+2^a+\cdots+n^a\right)^2\lt 1^{2a+1}+2^{2a+1}+\cdots+n^{2a+1}.$$

Proof for Lemma : Let $L(n), R(n)$ be the left hand side and the right hand side respectively. Note that $L(1)=R(1)$ for any $a$. By the way, since $$L(n+1)-L(n)=(n+1)^a\left\{(n+1)^a+2\sum_{k=1}^{n}k^a\right\},R(n+1)-R(n)=(n+1)^{2a+1},$$ we get $$\{R(n+1)-R(n)\}-\{L(n+1)-L(n)\}=(n+1)^{a}\left\{(n+1)^{a+1}-(n+1)^a-2\sum_{k=1}^nk^a\right\}$$ $$=(n+1)^a\left\{n(n+1)^a-2\sum_{k=1}^nk^a\right\}=(n+1)^{2a}\left\{n-2\sum_{k=1}^n\left(\frac{k}{n+1}\right)^a\right\}.$$ Here, since $a\ge 2$, we get $$2\sum_{k=1}^n\left(\frac{k}{n+1}\right)^a=\sum_{k=1}^n\left\{\left(\frac{k}{n+1}\right)^a+\left(\frac{n+1-k}{n+1}\right)^a\right\}\lt\sum_{k=1}^n1=n.$$ Hence, we know $$\{R(n+1)-R(n)\}-\{L(n+1)-L(n)\}\gt0\iff R(n+1)-L(n+1)\gt R(n)-L(n).$$ With $L(1)=R(1)$, the proof for lemma is completed.

Proof for Theorem : Supposing $c\le ab$, since $b\ge 2$, we get $$(1+2^a+\cdots+n^a)^b=\left(1+2^{ab}+3^{ab}+\cdots+n^{ab}\right)+\cdots\gt 1+2^c+\cdots+n^c.$$ This is a contradiction. Hence, $c\gt ab$.

Supposing $b\ge 3$, we get $c\gt ab\ge 3$. For $n=8k-5,8k-4\ (k\in\mathbb N)$, considering in mod $8$, since $1+2^a+\cdots+n^a$ is even, we have $$(1+2^a+\cdots+n^a)^b\equiv 0.$$ However, the right hand side of $(\star)$ is not a multiple of $8$. Let us prove this.

Note that $1^{2m-1}\equiv 1, 3^{2m-1}\equiv 3, 5^{2m-1}\equiv 5, 7^{2m-1}\equiv 7$ and that $1^{2m}\equiv 3^{2m}\equiv 5^{2m}\equiv 7^{2m}\equiv 1$ (mod $8$) for any $m\in\mathbb N$.

$(1)$ The $c=2m-1$ case : Noting that $(\text{even})^c\equiv 0$ for $c\ge 3$,
$$RHS\equiv 1+3+5+\cdots+(8k-5)=\sum_{i=1}^{4k-2}(2i-1)\equiv 4\not\equiv 0.$$

$(2)$ The $c=2m$ case : We have
$$RHS\equiv \sum_{k=1}^{4k-2}1=4k-2\not \equiv 0.$$ $(1)(2)$ tells us that the right hand side of $(\star)$ is not a multiple of $8$ for any $c$. This is a contradiction. Hence, we know $b=2, c\gt 2a$.

Supposing $a\ge 3$, since $$\left(\frac 1n\right)^a+\left(\frac 2n\right)^a+\cdots+\left(\frac{n-1}{n}\right)^a\le\left(\frac 1n\right)^3+\left(\frac 2n\right)^3+\cdots+\left(\frac{n-1}{n}\right)^3=\frac{(n-1)^2}{4n},$$ we get $$1^a+2^a+\cdots+(n-1)^a\le\frac{(n-1)^2}{4n}\times n^a.$$ Hence, $$n^c\lt 1+2^c+3^c+\cdots+n^c=(1+2^a+3^a+\cdots+n^a)^2\le \left(\frac{(n-1)^2}{4n}\times n^a+n^a\right)^2=n^{2a}\left(\frac{(n-1)^2}{4n}+1\right)^2=n^{2a}\left(\frac{n^2+2n+1}{4n}\right)^2\lt n^{2a+2}.$$ This is because $\left(\frac{n^2+2n+1}{4n}\right)^2\lt n^2\iff \frac{n^2+2n+1}{4n}\lt n\iff (n-1)(3n+1)\gt0\iff n\ge 2$. Hence, $n^c\lt n^{2a+2}$ leads $0\lt c-2a\lt 2$, which means that $c-2a=1\iff c=2a+1$. However, the above lemma tells us that this case has no solution. Hence, we know $a\le 2$.

The $(a,b)=(1,2)$ case leads $c=3$.

The $(a,b)=(2,2)$ case leads $(1+2^2+\cdots+n^2)^2=\left(\frac{n(n+1)(2n+1)}{6}\right)^2=1+2^c+\cdots+n^c$. However, since we can easily prove that $$\left(\frac{n(n+1)(2n+1)}{6}\right)^2\lt 1^5+2^5+\cdots+n^5=\frac{n^2(n+1)^2(2n^2+2n-1)}{12}$$ for $n\ge 2$, this case leads a contradiction.

Now, the proof for Theorem is completed.

PS: This idea (using mod $8$) does not seem to work for the other $n$. Another idea would be needed.