Does $f(x) \in \mathbb{Z}[x]$ irreducible, imply $f(2x)$ also irreducible?

It makes a difference if you are considering irreducibility in $\mathbb{Z}[x]$ or in $\mathbb{Q}[x]$.

A counterexample for the first case: $f(x) = x$ is irreducible in $\mathbb{Z}[x]$, but $f(2x) = 2\cdot x$ is not.

In the second case your statement is true for all $n\neq 0$, since then $n$ is a unit in $\mathbb{Q}[X]$, so $f(x)\mapsto f(nx)$ is an automorphism of $\mathbb{Q}[x]$.

UPDATE

After giving this answer,the following was added to the question

Edit: by "irreducible" I meant that f cannot be written as a product of two nonconstant polynomials.

First of all, this is a bit unfortunate, since it differs from the established definition of irreducibility in ring theory.

However by Gauss' Lemma, it turns out that your notion of "irreducibility" of a polynomial $f\in \mathbb Z[x]$ is exactly the same as the irreducibility of $f$ in $\mathbb Q[x]$, which is covered by the above discussion.