Dirichlet's theorem/Bunyakovsky conjecture for infinite composites/a single prime

Question

From https://en.wikipedia.org/wiki/Bunyakovsky_conjecture, the Bunyakovsky Conjecture is an open problem that states that $f(x)$ has infinitely many primes in sequence $f(1),f(2),...$ if

1) The leading coefficient is positive.

2) The polynomial is irreducible over the integers.

3) The coefficients of $f(x)$ are relatively prime.

It is conjectured as a generalization of Dirichlet's theorem on arithmetic progressions, which states that $a+xd$ contains infinitely many primes for $a$ and $d$ relatively prime.

I am curious whether it is known that $f(x)$ must contain at least one prime if the three conditions are met, or whether we know that $f(x)$ must have infinitely many composites for certain conditions, even just in the case of first degree polynomials.

Answer 1

It would appear that this is actually equivalent to Bunyakovsky's Conjecture. See Theorem 2 and the proof in "Prime values of irreducible polynomials" by Micheal Filaseta. This states:

Let $g$ be a positive integer. For any f be an irreducible polynomial of degree $g$, define $N_f = \text{gcd}\left\{ f(n) : n \in \{1, \ldots, g - 1\}\right\}$. If for each irreducible polynomial $f$ of degree $g$, there exists one integer $m$ for which $f(m)/N_f$ is prime, then for each irreducible polynomial $f$ of degree $g$, there exists infinitely many integers $m$ for which $f(m)/N_f$ is prime.

Here Micheal means both positive and negative primes, which is why they have dropped the condition that the leading coefficient is positive. You can find this paper here at http://matwbn.icm.edu.pl/ksiazki/aa/aa50/aa5024.pdf.