Generalizing the sum of consecutive cubes ∑nk=1k3=(∑nk=1k)2\sum_{k=1}^n k^3 = \Big(\sum_{k=1}^n k\Big)^2 to other odd powers

This is a partial answer, it just establishes the existence.

We have
$$s_m(n) = \sum_{k=1}^n k^m = \frac{1}{m+1} \left(\operatorname{B}_{m+1}(n+1)-\operatorname{B}_{m+1}(1)\right)$$
where $\operatorname{B}_m(x)$ denotes the monic
Bernoulli polynomial
of degree $m$, which has the following useful properties:
$$\begin{align} \int_x^{x+1}\operatorname{B}_m(t)\,\mathrm{d}t &= x^m \quad\text{(from which everything else follows)} \\\operatorname{B}_{m+1}'(x) &= (m+1)\operatorname{B}_m(x) \\\operatorname{B}_m\left(x+\frac{1}{2}\right) &\begin{cases} \text{is even in $x$} & \text{for even $m$} \\ \text{is odd in $x$} & \text{for odd $m$} \end{cases} \\\operatorname{B}_m(0) = \operatorname{B}_m(1) &= 0 \quad\text{for odd $m\geq3$} \end{align}$$

Therefore,
$$\begin{align} s_m(n) &\text{has degree $m+1$ in $n$} \\s_m(0) &= 0 \\s_m'(0) &= \operatorname{B}_m(1) = 0\quad\text{for odd $m\geq3$} \\&\quad\text{(This makes $n=0$ a double zero of $s_m(n)$ for odd $m\geq3$)} \\s_m\left(x-\frac{1}{2}\right) &\begin{cases} \text{is even in $x$} & \text{for odd $m$} \\ \text{is odd in $x$} & \text{for even $m\geq2$} \end{cases} \end{align}$$

Consider the vector space $V_m$ of univariate polynomials $\in\mathbb{Q}[x]$
with degree not exceeding $2m+2$, that are even in $x$ and have a double zero
at $x=\frac{1}{2}$.
Thus $V_m$ has dimension $m$ and is clearly spanned by
$$\left\{s_j^2\left(x-\frac{1}{2}\right)\mid j=1,\ldots,m\right\}$$
For $m>0$, we find that $s_{2m+1}(x-\frac{1}{2})$ has all the properties
required for membership in $V_m$.
Substituting $x-\frac{1}{2}=n$, we conclude that there exists a representation
$$s_{2m+1}(n) = \sum_{j=1}^m a_{m,j}\,s_j^2(n) \quad\text{for $m>0$ with $a_{m,j}\in\mathbb{Q}$}$$
of $s_{2m+1}(n)$ as a linear combination of squares of sums.