No polynomial takes only prime values

A value of absolute value at least $2$. A non-constant polynomial is unbounded, so there is an integer $n_0$ with $|P(n_0)|\ge2$. Put $m=P(n_0)$.

Divisibility. For integer polynomials, $a-b\mid P(a)-P(b)$. With $a=n_0+km$, $b=n_0$ we get $m\mid P(n_0+km)-m$, so $m\mid P(n_0+km)$ for every integer $k$.

Large values. A non-constant polynomial takes each value at most $\deg P$ times, so for all but finitely many $k$, $P(n_0+km)\notin\{m,-m\}$. Taking also $|n_0+km|$ large enough that $|P(n_0+km)|>|m|$, we get a multiple of $m$ whose absolute value is strictly larger than $|m|\ge2$.

Conclusion. Such a number $|P(n_0+km)|$ has the divisor $|m|$ with $1<|m|<|P(n_0+km)|$, so it is composite. This happens for infinitely many $k$, giving infinitely many integers $n=n_0+km$. $\blacksquare$

Remark. Famous examples such as $n^2+n+41$ are prime for many consecutive $n$ (here $n=0,\dots,39$), but must fail eventually (at $n=41$, $P(41)=41\cdot43$).

Lesson. The congruence $a-b\mid P(a)-P(b)$ lets you make a polynomial value divisible by a number you control.