Regarding identities with sums of consecutive squares

This comic http://abstrusegoose.com/63 points out an interesting identity with sums of consecutive squares. Let us take positive integers $k, p, q$, with $p < q$, and ask if $k^2 + \ldots + (k + p)^2 = (k + p + 1)^2 + \ldots + (k + q)^2$. One may see this is equivalent to $(p+1)(6k^2 + 6kp + 2p^2 + p) = (q – p)(6k^2 + 6k(p + q + 1) + 2p^2 + p(2q + 3) + 2q^2 + 3q + 1)$. Surprisingly, there is an easy family of triples $(k, p, q)$ of solutions: $(p(2p + 1), p, 2p)$. Searching a bit, I found a few values of $k,p,q$ which solve the equation and are not of the given form:

4 34 44

12 38 51

16 126 163

18 16 24

60 50 75

67 92 131

What can we say about the solutions which are not of that form? Are there infinitely many? Is there another form?

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