To make things harder, we prove a result about polynomials that have shape very much like the one described by the OP.
We will show that there are infinitely many irreducible polynomials $x^{2011} +p(p+2)x+p$ with $p$ prime and $x^{2011} +p(p+2)x+p+2$ irreducible. For each of these two polynomials, the irreducibility proof uses the Eisenstein Criterion.
A possible approach is to show that $p+2$ is prime for infinitely many primes $p$. Then, as in the OP's example, $x^{2011} +p(p+2)x+p$ and
$x^{2011} +p(p+2)x+p+2$ are both irreducible by Eisenstein's Criterion. However, this requires proving the Twin Prime Conjecture, so it is not the easy way to go.
Instead, recall that a number $n$ is called powerful(l) if for every prime $q$ that divides $n$, its square $q^2$ divides $n$. Powerful numbers are relatively scarce. There is a constant $C$ such that the number of powerful numbers less than $x$ is less than $Cx^{1/2}$ (see for example the Wikipedia article).
Since the primes have much greater asymptotic density than the powerful numbers, there are infinitely many primes $p$ such that $p+2$ is not powerful. Indeed, for "most" primes $p$, the number $p+2$ is not powerful. Let $p$ be any prime such that $p+2$ is not a powerful number.
Then $x^{2011}+p(p+2)+p$ and $x^{2011}+p(p+2)+p+2$ are both irreducible. For the second polynomial, we use the Eisenstein Criterion with $q$ any prime which divides $p+2$ but whose square does not divide $p+2$. There is such a prime $q$ since $p+2$ is not powerful.
Comment: The proof in principle is of irreducibility over $\mathbb{Q}[x]$. But irreducibility over $\mathbb{Z}[X]$ is an immediate consequence, since our polynomial is monic, and consequently the $\gcd$ of the coefficients is $1$.
By using Chen primes, we can find infinitely many primes $p$ such that $p+2$ is prime or has at most two prime factors. That is the nearest we can come to imitating your $3$, $5$ example.