If $f\in\mathbb Z[X]$ and there are $x_1,\dots, x_n\in\mathbb Z_+$, where $n>\deg f$, such that $f(x_i)\in\mathbb P$, $i=1,\dots n$, then $f$ is irreducible over $\mathbb Z$. Because, if $\,f=g\cdot h$ then either $g(x_i)=1$ or $h(x_i)=1$ which can't happen $n>\deg g+\deg h$ times.
But what about the opposite? Are there non constant, irreducible polynomials $p\in\mathbb Z[X]$ such that
$x_1,\dots, x_n\in\mathbb Z_+$ and $p(x_i)\in\mathbb P$ for all $i=1,\dots,n$ $\:\implies\:$ $n\leq \deg p$?
Your claim seems to be negative.
For example, if we take $f(x)=x^3-x+3$. It is an irreducible polynomial of degree $3$, and for each positive integer $n$, $f(n)$ is divided by $3$ ( By Fermat's little theorem). Since $f(n)>3$ when $n>2$, the only $n$ such that $f(n)\in \mathbb{P}$ is just $1$.