Sets of Prime Numbers Generated By an Irreducible Monic Polynomial Given a non-constant integral irreducible monic polynomial $f(x)$, the prime factors of its value at integers $x\in\mathbb{N}$ forms a set $\mathcal{P}(f)$.

Is it possible that $\mathcal{P}(f)\cap\mathcal{P}(g)=\emptyset$ for two such polynomials $f$ and $g$.
The answer is no. See Theorem 3 in Prime Numbers in Certain Arithmetic Progressions:


  If $f,g \in \mathbb{Z}[x]$ are non-constant, then $P(f) \cap P(g)$ is infinite.


The proof there is quite short so I might as well include it:


  Suppose $\alpha$ is a root of $f$ and $\beta$ is a root of $g$. By Dedekind 's theorem, except for a finite number of exceptions, $p$ is a prime divisor of $f$ exactly when $p$ has a first degree prime ideal factor in the field $\mathbb{Q}(\alpha)$. A similar statement holds for $g$. Consider the field, $\mathbb{Q}(\alpha, \beta)=\mathbb{Q}(\theta)$ for some $\theta \in \mathcal{O}_K$. If $h$ is the minimal polynomial of $\theta$ then Schur's Theorem assures us that there are infinitely many primes which have a first degree prime ideal factor in $\mathbb{Q}(\alpha, \beta)$. These primes thus have a first degree prime ideal factor in both $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\beta)$. Thus, except for a finite number of exceptions, these primes are in both $P(f)$ and $P(g)$.
Let $f(y)=y^2-2$, and let $g(x)=x^3+5$, then this works, because if they had a solution in common you'd have a solution for $y^2=x^3+7$, but this is known not to have any integer solutions. See Silverman's Arithmetic of Elliptic curves page 296.