Bound the product of the 2-full parts of $k$ consecutive integers: is it $n^{2+o(1)}$? (Erdős #367)

For a positive integer $m$, write $B_2(m) = m/m'$, where $m'$ is the product of the primes that divide $m$ to the first power exactly; equivalently $B_2(m) = \prod_{p^a\parallel m,\ a\geq 2} p^a$ is the *powerful ($2$-full) part* of $m$. Erdős and Graham ask: is it true that for every fixed integer $k\geq 1$, $$\prod_{n\leq m < n+k} B_2(m) \ll n^{2+o(1)}\,?$$ A sharper form asks whether even $\prod_{n\leq m<n+k} B_2(m) \ll_k n^2$. More generally, for the *$r$-full part* $B_r(m) = \prod_{p^a\parallel m,\ a\geq r} p^a$, one asks whether, for every fixed $r,k\geq 2$ and $\epsilon>0$, $$\limsup_{n\to\infty}\frac{\prod_{n\leq m<n+k} B_r(m)}{n^{1+\epsilon}} = \infty.$$

Acceptance. FULLY RESOLVES: a complete proof (machine-checkable in Lean/Coq preferred, else a full written proof) settling, for every fixed $k\geq 1$, whether $\prod_{n\leq m<n+k}B_2(m)\ll n^{2+o(1)}$ — either establishing the bound or exhibiting a $k$ and an infinite family of $n$ that violate it. A complete resolution of the $r$-full generalisation ($\limsup_n \prod_{n\leq m<n+k}B_r(m)/n^{1+\epsilon}=\infty$ for all fixed $r,k\geq 2$, $\epsilon>0$) fully resolves that variant. ADVANCES (each independently checkable): a proof of the $n^{2+o(1)}$ bound for a specific fixed $k\geq 3$ (the first cases beyond the trivial $k\leq 2$); OR a strictly better unconditional upper bound than the trivial $\prod_{n\leq m<n+k}B_2(m)\ll n^{k}$ for some fixed $k\geq 3$, with proof; OR a reproducible computation exhibiting products of size $\gg n^{2}g(n)$ for an explicit slowly-growing $g$ that sharpens van Doorn's $n^2\log n$ lower bound, together with the search code and certificates. Deliver the proof file or the search code plus the record configurations and bounds.