For which $k\ge 2$ does $(n+k)!^2\mid(2n)!$ hold for infinitely many $n$? Search the divisibility (Erdős #727)

Fix an integer $k\ge 2$. Erdős, Graham, Ruzsa and Straus asked whether $(n+k)!^2\mid(2n)!$ holds for infinitely many positive integers $n$. For any given $n$ and $k$ this divisibility is decidable by prime valuations: $(n+k)!^2\mid(2n)!$ iff for every prime $p\le 2n$ one has $v_p((2n)!)\ge 2\,v_p((n+k)!)$, where $v_p(m!)=\sum_{i\ge1}\lfloor m/p^i\rfloor$ (Legendre). GOAL (computational): for each small $k\ge 2$, enumerate all $n\le N$ satisfying the divisibility, tabulate the solution sets, and report whether solutions appear to be finite or to recur (e.g. supported on an arithmetic-progression / smooth-$n$ pattern) as $N$ grows.

Acceptance. PARTIAL / PRIMARY (finite, machine-checkable): for each of several values $k=2,3,\ldots$, the exact set $\{n\le N:(n+k)!^2\mid(2n)!\}$ (state $N$), computed via Legendre-valuation divisibility (no huge-integer arithmetic needed), delivered with a runnable script and the tabulated solution counts vs. $N$. A clear empirical dichotomy per $k$ (bounded solution set vs. steady recurrence) is the target evidence, cross-checked against A002503/A343507/A389396 where they apply. FULLY RESOLVES: a proof, for some $k\ge2$, that the solution set is infinite (or finite). State $N$ and the range of $k$ covered.