Let $P\in{\mathbb Z}[X]$ be a nonconstant polynomial. For $t\in{\mathbb N}$, let $P_t(X)=P(X)-t$.
Suppose that $P$ can be decomposed as $P=A(B(X))$ for some nonconstant $A,B\in{\mathbb Z}[X]$ and that $t=A(u)$ for some $u\in{\mathbb Z}$. In that situation, $P_t$ is obviously not irreducible, as it is divisible by $B(X)-u$.
My question is about the converse : is there a constant $M$ such that whenever $t\geq M$ and there are no $A,B,u$ as above, then $P_t$ is irreducible.
My thoughts : this is true when $P=X^n$ (we can take $M=2$ in this case).
Edit by @MooS:
As there were some misinterpretations, let me rephrase the question:
Let $P \in \mathbb Z[X]$ and $t \in \mathbb N$. We call the pair $(P,t)$ strongly reducible if there are polynomials $A,B \in \mathbb Z[X], u \in \mathbb Z$ with $1 \leq \deg B < \deg P$, such that $P=A(B(x))$ and $t=A(u)$.
Any strongly reducible pair clearly yields the reducible polynomial $P-t$, since $P-t$ is divisible by the lower degree polynomial $B-u$.
Now there are a few questions to investigate:
Does every reducible polynomial $F$ come from a strongly reducible pair $(F+t,t)$ for some $t \in \mathbb N$? This is false.
Given a pair $(P,t)$ such that $P-t$ is reducible, is $(P,t)$ strongly reducible? This is false.
Since 2. is false, are there only finitely many exceptions for a fixed $P$ and varying $t$? I.e. do almost all reducible $P-t$ come from a strongly reduced pair $(P,t)$? I think this is precisely what the OP was asking.