With motivation from the original poster, a key idea from Sylvester, and technical inspiration from Iosif Pinelis, I contribute an observation that helps toward an answer.

I use m instead of n and n instead of k. I start with the inequality that p! is strictly less than 3^p for p less than 7, and less than (p/2)^p for all larger integers p. We will set p =$\pi$(n).

Consider the product of integers in (m,m+n], and write it as W(n!)B, where W are all the prime factors (with multiplicity) at most n dividing (m+n)!/(m!n!), leaving B as the product of the remaining prime factors which are all larger than n, and B=1 if there are no such large prime factors.

Sylvester's observation is that W is at most (m+n-p+1)...(m+n). If B=1 the interval (m,m+n] has all numbers being n-smooth. The extended observation (which I think is new and hopefully orignal) is that WB is at most (m+n-p-d+1)...(m+n) if there are at most d many numbers in (m,m+n] which are not n-smooth. We fix d and observe that the original problem relates to d=1 in what follows.

Under supposition that there are not d+1 many non smooth numbers in (m,m+n], we now have (n!) Is at least (m+1)...(m+n-p-d). Write m as kn + i for positive integer k and non negative integer i (choose i less than n for less confusion). We now have (p+d)! Is at least (and for large enough n strictly greater than) k^(n-p-d).

So if (m,m+n] has at most d numbers which are not n-smooth, then we use the inequality above to note that when p+d is greater than 6, k is strictly less than ((p+d)/2)^((p+d)/(n-(p+d))). To save on plus signs, write q=p+d.

By the above, when q is at most 6 and n is at least 2q, then k is at most 2. (I leave the case n smaller than 12 and arbitrary d to the reader.) As n grows, q(1+ log (q/2)) will be less than n (because d is fixed), and one can use current literature or supercomputers to compute for which n this holds, in which case k is strictly less than e.

So given d, one can compute n0 without much challenge to find that (m,m+n] has d+1 non smooth numbers for n greater than n0 and for m at least as large as 3n.

To handle the remaining case for small d (d less than 6), use Nagura or similar as outlined in the other answer of mine to find d+1 non smooth integers in the interval for when m is in [n,3n). This should hold for m at least 150, giving C is less than 150.

Gerhard "Would James Joseph Be Approving?" Paseman, 2020.06.01.