Something wrong my proof that $\mathbb{Z}[x]$ is not a PID nor an Euclidean domain?

Question

My proof goes as follows:

Suppose for contradiction that $\mathbb{Z}[x]$ is a PID. Then the ideal generated by any irreducible element is maximal. We know that $x^2+1$ is irreducible in $\mathbb{Z}[x]$ so $(x^2+1)$ should be maximal; $(x^2+1) \ne \mathbb{Z}[x]$ since $1 \notin (x^2+1)$. However, $(x^2+1) \subset (x^2+1, x)$, and the containment is strict since $x \notin (x^2+1)$. Thus $(x^2+1)$ is not maximal, and $\mathbb{Z}[x]$ is not a PID, and therefore also not an Euclidean domain.

I feel like I made a mistake somewhere because $x^2+1$ is also irreducible over $\mathbb{Q}$ so the same argument should work for $\mathbb{Q}[x]$, but I know that $\mathbb{Q}[x]$ is an Euclidean domain (and thus a PID) since $\mathbb{Q}$ is a field. Where is my mistake? Am I wrong to say $(x^2+1) \subset (x^2+1, x)$ when considered as ideals over $\mathbb{Q}[x]$?

Answer 1

The example does not work, because the ideal generated by $x^2+1$ and $x$ is all of the ring.

Try the ideal generated by $2$ and $x$. It should not be hard to show that this ideal is not principal.

Answer 2

Well, "the ideal generated by any irreducible element is maximal" is not true.
Instead, "the ideal generated by any irreducible element is prime" is always true.

In fact, irreducibility means "$fg∈I \Rightarrow f∈I$ or $g∈I$", which is the property of prime ideal.

EDIT I was also wrong: A counterexample is $3 \in Z[\sqrt{-5}]$ (referred to in Wikipedia).
The truth is "the ideal generated by any prime element is prime".