Cohn Irreducibility:
We can use a generalization of Cohn's Irreducibility Criterion to get a bound.
The main idea behind the proof of the Cohn criterion is to show that for any polynomial $f(x)$ fulfilling the criterion for a given base $b$, every root $\alpha$ of $f(x)$ satisfies $|b-\alpha| > 1$. (Except for the case $b=2$, which involves an additional argument.)
From there consider any nontrivial factor $f_1(x)$ of $f(x)$, and write $f_1$ in the form
$$f_1(x) = c\prod_i(b-\alpha_i)$$ for some integer $c$ and subset $\{\alpha_i\}$ of the roots of $f$. But then
$$|f_1(b)| = |c|\prod_i|b-\alpha_i| > 1.$$
This would contradict the fact that at least one of the factors of $f(b)$ is $\pm 1$ if $f(b)$ is prime.
To show that $|b-\alpha| > 1$, M. Ram Murty's proof of Cohn's Irreducibility Criterion contains the following lemma:
Lemma 2. Let $f(x) = a_nx^n + a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ belong to $\mathbb{Z}[x]$. Suppose that $a_n \ge 1, a_{n-1}\ge 0$, and $|a_i| \le H$ for $i=0,1,\ldots,n-2$, where $H$ is some positive constant. Then any complex zero $\alpha$ of $f(x)$ either has nonpositive real part or satisfies $$|\alpha| < \frac{1+\sqrt{1+4H}}{2}.$$
To get Cohn's Irreducibility Criterion we use $H=b-1$ in the lemma to get
$$\Re(\alpha) \le 0 \qquad \text{ or } \qquad |\alpha| < \frac{1+\sqrt{1+4(b-1)}}{2} < b-1 \tag{1}$$
for $b\ge 3$, and in either case we get $|b-\alpha| > 1.$
However, the second inequality in $(1)$ is very crude. If instead we use (for instance)
$$\frac{1+\sqrt{1+4(b-1)}}{2} < \sqrt{b} + \tfrac{1}{2},$$
then we get
$$|b - \alpha| \ge b - |\alpha| > b - \sqrt{b}-\tfrac{1}{2}.$$
Using the same argument as above, this results in the following generalization of the Cohn Irreducibility Criterion:
Theorem. Let $f(x) = a_nx^n + a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ belong to $\mathbb{Z}[x]$. Suppose that $a_n \ge 1, a_{n-1}\ge 0$, and $|a_i| \le b-1$ for $i=0,1,\ldots,n-2$, where $b$ is an integer larger than $2$. If $f(x)$ is reducible, then $f(b)$ can be factored into two positive integers $p$ and $q$ such that
$$b - \sqrt{b} - \tfrac{1}{2} < p \le q.$$
For your specific question, any $b$ that is large enough to satisfy
$$ g < b - \sqrt{b}-\tfrac{1}{2} $$
will work.
Bunyakovsky Conjecture:
Yes, if we assume the Bunyakovsky conjecture, then for any irreducible polynomial $f(x)$ in $\mathbb{Z}[x]$ with positive leading coefficient and $g := \gcd(f(1), f(2), f(3),\ldots)$, then $\frac{f(b)}{g}$ will be prime for infinitely many values of $b \in \mathbb{N}$.
Proof: (sketch)
Find an $r$ such that $\frac{1}{g}f(r)$ is relatively prime to $g$. (In other words, $\gcd(f(r), g^2) = g$. You can use the Chinese Remainder Theorem to construct such an $r$.)
I claim that the polynomial $h(x) = \frac{1}{g}f(g^2x+r) \in \mathbb{Z}[x]$ satisfies the conditions of the Bunyakovsky conjecture:
- $h$ has a positive leading coefficient, since $f$ does.
- $h$ is irreducible over $\mathbb{Q}$, since $f$ is. (Strictly speaking the conjecture states $h$ must be irreducible over $\mathbb{Z}$, but this follows from irreducibility over $\mathbb{Q}$ together with the next condition.)
- $\gcd(h(1), h(2), h(3),\ldots) = 1.$
To prove 3, we know that for any integer $n$,
$$f(g^2n+r) \equiv f(r) \pmod{g^2}$$
so
$$h(n) = \frac{1}{g}f(g^2n+r) \equiv \frac{1}{g}f(r) \pmod{g}.$$
So for any prime $p$ dividing $g$,
$$h(n) \equiv \frac{1}{g}f(r) \not\equiv 0 \pmod{p},$$
hence $p$ can't divide $\gcd(h(1), h(2), h(3),\ldots)$.
And for any prime $q$ not dividing $g$, the map $n \mapsto g^2n+r$ permutes the integers mod $q$, so modulo $q$ the set $\{f(n)\}$ contains the same values as the set $\{f(g^2n+r)\}$. From the definition of $g$, $q$ does not divide all of the $f(n)$; so $q$ does not divide all of the $f(g^2n+r)$; so $q$ does not divide all of the $h(n)$. Hence $q$ doesn't divide $\gcd(h(1), h(2), h(3), \ldots)$.
So no primes divide $\gcd(h(1), h(2), h(3), \ldots)$, and the result follows.
Best Answer
Try this for the polynomial $2X^2 + 5X + 2$.