How many times can a $4^{th}$ degree polynomial be equal to a prime number?

It is an open problem. In fact, by the Dirichlet's theorem, linear polynomial $f(x)=ax+b$ takes infinitely many prime values, provided $(a,b)=1.$ For all other polynomials of degree $> 1$ it is not known whether there exists a polynomial which takes infinitely many primes. The conjecture is that even $f(x)=x^2+1$ can take infinitely many prime values. If you allow yourself to go to other higher dimensions, then $f(x,y)=x^2+y^2$ would take all prime valuse of the form $4k+1$ by the theorem of Fermat.

This is most likely an open problem.

There are two cases:

• $f$ is reducible in $\mathbb Z[x]$: Then $f=gh$ for non-constant polynomials $g,h\in\mathbb Z[x]$. If $f(x)$ is a prime, then either $g(x)=1$ or $h(x)=1$. $g-1$ is not the zero polynomial, so it has at most $\deg g$ roots, so $g(x)=1$ for at most $\deg g$ many $x$. Same for $h$, so there are at most $\deg g+\deg h=\deg f$ values $x\in\mathbb Z$, such that $|f(x)|$ is prime.

• $f$ is irreducible: The Bunyakovsky conjecture states, that $|f(x)|$ is prime for infinitely many $x\in\mathbb Z$, where $f$ is an arbitrary irreducible polynomial. For $\deg f=1$, this was in fact proven by Dirichlet, for $\deg f=4$, it seems to be unsolved.