Find a pair $(n, k)$ such that $\sum_{i=1}^{n} i = \sum_{i=1}^{k} i^2$?

There are only two variables involved. If you want to search, you can write it as a quadratic in $n$, just try values of $k$, solve for $n$, and see if it comes out integral. I find k=5, n=10, k=6, n=13 and k=85, n=645 as solutions as well with no more under k=200. Then OEIS has no more and asserts the series is finite. There are references for this claim in A053611

Fix the variable $k$. Let $$k' = \dfrac{k(k+1)(2k+1)}{6}.$$ Then you get the quadratic equation $$n^2+n-2k' = 0$$ with the solutions $$n_{1/2} = -\dfrac{1}{2} \pm \sqrt{(\dfrac{1}{2})^2+2k'}.$$ Now you can generate your solution pairs.