On the no trivial $3$-tuples $(p, q, \alpha) \in \mathbb{N}^3$ such that $\sum_{k = 1}^{n}k^p =\Big [\sum_{k=1}^{n}k^q\Big]^\alpha $. It is well known that $\sum_{k = 1}^{n}k^3 =\Big [\sum_{k=1}^{n}k^1\Big]^2$.  My question is very simple.

There are   $3$-tuples $(p, q, \alpha) \in
 \mathbb{N}\times\mathbb{N}\times\mathbb{N}$, in addition to $(3,1,2)$,
   such that $\alpha\geq 2$ and  $$\sum_{k = 1}^{n}k^{\,p} =\Big [\sum_{k=1}^{n}k^{\,q}\Big]^\alpha, \quad \forall n \in \mathbb{N}{\;\Large ?}$$    

 A: [The following claims should be obvious, and will not be proven.]
Claim: The degree of $ \sum_{i=1}^n i^r$ is $r+1$.
Claim: The leading coefficient of $ \sum_{i=1}^n i^r$ is $\frac{1}{r+1} $.
The first claim gives us $p+1 = \alpha (q+1) $. The second claim gives us $\frac{1}{p+1} = \left( \frac{ 1}{q+1} \right)^\alpha$.
Multiplying these two equations, we get $(q+1) = \alpha^\frac{1}{\alpha-1}$
Claim: $1 \leq \alpha^{\frac{1}{\alpha-1}} \leq 2$, with equality if and only if $\alpha = 1, 2$. 
With this claim, the RHS is never an integer except for $\alpha = 1, 2 $.
If $\alpha = 1$, then $p+1 = q+1, p=q$.
If $\alpha = 2$, this forces $q+1=2$ or $q=1$, and that $p+1 = 2(q+1) $ so $p=3$.
Finally, verify that $(p,p,1) $ and $(3,1,2)$ are solutions.
A: Let $n=2$, so $(1+2^p)=(1+2^q)^{\alpha}$, so $(1+2^q)^{\alpha}-2^p=1$. If $p=1$, then $q=\alpha=1$, contradicting $\alpha \geq 2$. Otherwise $1+2^q, \alpha, 2, p>1$, so by Mihailescu's theorem $1+2^q=3, \alpha=2, p=3$. This gives $(3, 1, 2)$, which is indeed a solution. 
