This is false as stated. Take, for example, $p(n) = n^2 + n$. The correct condition is that $p$ is irreducible and there exists no $d > 1$ such that $d | p(n)$ for all $n$. With this condition this is a big open problem, the Bunyakovsky conjecture, and it is open for any such polynomial $p$ of degree greater than $1$.
This is false: The polynomial $X^2+X+2$ is irreducible, has positive degree and its coefficients have no common factors, but, since $n,n^2$ always have the same parity, it always takes even values, so the only prime value it can take is $2$. Also, consider polynomial such as $-(X^2+1)$: it always takes negative values, so it can't have prime values.
However, it is conjectured that these are the only possible kinds of counterexample. Indeed, Bunyakovsky conjecture states that the polynomial $P$ in your question takes infinitely many prime values if it satisfies two additional constraints:
- $P$ has a positive leading coefficient (so that, for large $n$, $P(n)>0$), and
- there is no prime $p$ which divides $P(n)$ for all $n\in\mathbb Z$.
This conjecture has not, however, been established in any nonlinear case. That is, there is no known polynomial of degree $>1$ which takes infinitely many prime values. An example here is the polynomial $P(X)=X^2+1$, which is one of the four famous Landau's problems.
By the way, just for fun, let me mention the Bateman-Horn conjecture. This is a wild generalization of the Bunyakovsky conjecture, which, first, concerns not just prime values of single polynomials, but multiple polynomials simultaneously taking prime values, and not only states there are infinitely many of these, but also gives conjectured asymptotic number of such values. It is somewhat impressive how far people can come with generalizing problems for which we have no clue how to solve even the simplest of cases...
Best Answer
No, $x^2+2x+1=(x+1)^2$ obviously is not prime for natural $x>0$.
Edit: for irreducible case you may want to read about the (unsolved) Bunyakovsky conjecture.