For any integer polynomial $f(t) \in \mathbb{Z}[t]$ (with positive leading coefficient), Hardy-Littlewood's Conjecture F provides a conjectural estimate for the number of primes that $f$ assumes less than $n$. To be a bit more specific, we have
$\#\{k\leq n \mid f(k) \text{ is prime}\} \sim c \cdot \frac{n}{\log n}$
for some constant $c$ dependent on $f$. In particular, the density of the set $\{k \mid f(k) \text{ is prime}\}$ is zero. Is there a way to prove this last fact without appealing to Conjecture F?
Perhaps the following observation will be helpful: let $p \mid f(k)$ for some prime $p$ and integer $k$. Then $f$ admits a root in $\mathbb{F}_p[t]$, so $p \mid f(k+mp)$ for any integer $m$. As $f$ is increasing (eventually), it follows that $f(k+mp)$ is not prime, for $m \gg 0$. Thus $f$ is not prime on a set of density at least $1/p$. As membership in the sets $p_1\mathbb{N}$ and $p_2\mathbb{N}$ is asymptotically independent (for $p_i$ prime), it follows that $f$ is composite on a set of density
$1-\prod_{\exists n:\;p \mid f(n)} \left(1-\frac{1}{p}\right)$
So in this sense, $f$ having too many prime factors actually implies that $f$ has attains prime values on a set of smaller density.