As pointed out in the other answers, the point is that there is no single-variable polynomial with integer coefficients which takes on only prime values - apart from constant polynomials. My preferred proof is:
Suppose $f(x)$ is prime for all positive integer $x$. Take any value of $f$, say $p=f(1)$. The behavior of $f$ modulo $p$ is periodic, so since $f(1)$ is zero modulo $p$, there are infinitely many $x$ such that $f(x)$ is zero modulo $p$. But they can't all be $p$ itself, since a polynomial can't take on the same value infinitely many times (for a couple of reasons - first because of growth rates, and second because $f(x)-p$ would be a polynomial with infinitely many roots).
Note that the simple statement that $f(x)$ is not prime for all $x$ is not enough to claim that it's not prime for all $x$ above a certain bound, but if you understand how the above proof works it's not hard to see that that's true as well.