(This is only a comment.) If you instead look at the family $\{x^d-x+n\}$, then it is known that there are infinitely many irreducible elements (for each positive $n$). It suffices to show that $f(x)=x^p-x+n$ is irreducible in $\mathbb{F}_p$ for any prime $p$ such that $p\nmid n$. This is similar to a homework problem often assigned from Dummit and Foote's "Abstract Algebra" textbook, which is also solved by David Speyer here: Is $x^p-x+1$ always irreducible in $\mathbb{F}_p$? The proof with $1$ replaced by $n$ is unchanged.

I suspect that the example family you gave also is known to have a positive answer.

(This is only a comment.) If you instead look at the family $\{x^d-x+n\}$, then it is known that there are infinitely many irreducible elements (for each positive $n$). It suffices to show that $f(x)=x^p-x+n$ is irreducible in $\mathbb{F}_p$ for any prime $p$ such that $p\nmid n$. This is similar to a homework problem often assigned from Dummit and Foote's "Abstract Algebra" textbook, which is also solved by David Speyer here: Is $x^p-x+1$ always irreducible in $\mathbb{F}_p$? The proof with $1$ replaced by $n$ is unchanged.

I suspect that the example family you gave also is known to contain infinitely many irreducible elements.