Let us call a positive integer $k$ a square-pivot, if there is a pair of integers $m \gt 0$ and $n \ge k$, such that the sum of the $(m+1)$ consecutive squares up to $k$ equals the sum of the $m$ consecutive squares from $(n+1)$ on:

$$(k - m)^2 + \cdots + k^2 = (n + 1)^2 + \cdots + (n + m)^2.$$

Some small square-pivots are

$\mathbf 4$: $3^2 + \mathbf 4^2 = 5^2$
$\mathbf{21}$: $20^2 + \mathbf{21}^2 = 29^2$
$\mathbf{24}$: $21^2 + 22^2 + 23^2 + \mathbf{24}^2 = 25^2 + 26^2 + 27^2$
$\mathbf{110}$: $108^2 + 109^2 + \mathbf{110}^2 = 133^2 + 134^2$

Find the sum of all distinct square-pivots $\le 10^{10}$.