Let $f\in\mathbb Z[X]$, define $\operatorname{P}^+(f)$ as the number of primes $>0$ that $f$ assumes at distinct integral arguments.
Theorem: If $f\in\mathbb Z[X]$ is non constant and reducible of degree $n$, then $\operatorname{P}^+(f)\leq n$. And for all $n$ there are non constant reducible polynomials of degree n such that $\operatorname{P}^+(f)=n$.
[Acta Arith.,104.2 (2002) 117-127.]
Most polynomials are irreducible but using the theorem for an irreducibility test would be inefficient since a lot of irreducible polynomials has a fixed divisor $>1$ and wouldn't pass the test.
A conjecture related to the conjecture of Bunjakowsky states:
Conjecture: If $ f\in\mathbb Z[X]$ is non constant and irreducible, then $f(a)/d$ assumes primes for an infinit number of distinct integral arguments $a$, where $d$ is the largest fixed divisor of $f$.
Irreducibility of Polynomials Whose Coefficients are Integers Page 32.
This makes me wonder if the following hypothesis is true:
$f\in\mathbb Z[X]$ of degree $n>0$ with coprime coefficients is irreducible, iff $\;\operatorname{P}^+(d^{-1}\cdot f)> n$ or $\;\operatorname{P}^+(d^{-1}\cdot (-f))> n$, where $d$ is the greatest fixed divisor of $f$.
$\operatorname{P}^+$ is extended above and defined even for integer-valued polynomials with rational coefficients. Proofs or counter-examples?
With a test program using the hypothesis on Eisenstein polynomials $f$ with random coefficients between $-19$ and $19$ and random degree between $2$ and $5$, testing both $f$ and $-f$, resulted in no miss in $1,000,000$ polynomials. The only drawback is the risk of overflow when evaluating the polynomials for higher degrees and greater coefficients.
Let $f \in \mathbb{Z}[X]$. It is said irreducible iff $f(X) = g(X)h(X) \implies g(X)=\pm1$ or $h(X) = \pm1$. Let $d = gcd(f(\mathbb{Z}))$. Assume $d=1$ and $| f(n)|$ is prime more than $2 \deg(f)$ times.
If $f(X) = g(X)h(X)$ is reducible, then $f(n) = \pm p$ implies $n$ is a root of $g(X)^2-1$ or $h(X)^2-1$. But those polynomials are of degree $2\deg(g), 2 \deg(f)-2\deg(g)$, so they have at most $2\deg(f)$ roots. A contradiction. Whence $f$ is irreducible.
To improve the bound $2 \deg(f)$, you need to split each case : $g(n)=1,g(n)=-1,h(n)=1,h(n)=-1$ and factorize.
Then look at the case $d=2$, you'll have more cases and some of them will probably allow more than $\deg(f)$ prime values.