So I know the case with $\langle 2,x\rangle$which is not a principal ideal, however I've seen other questions with stuff like $\langle 5,x^2+3\rangle$ which is a principal ideal, however what I don't get is the fact that I can make the proof from $\langle 2,x\rangle$ to my case, and it would say it's not a principal ideal, but $x^2$ is irreducible over $\Bbb Z_5$ so you can say it's principal ideal.
I'm very bad at maths by the way, I'm just trying to pass an exam, help. Also, maybe a better question would be why can't we say that $x$ is irreducible over $\Bbb Z_2$ in the case of $\langle 2,x\rangle$ so that it would come to the conclusion it is a principal ideal
$\mathbb{Z}[X]/\langle 5,X\rangle \cong \mathbb{Z}_5$ is a field so $\langle 5,X\rangle$ is a prime ideal (it is even maximal). On the other hand, note that $\mathbb{Z}[X]/\langle 2, X^2\rangle \cong \mathbb{Z}_2[X]/(X^2)$ is not a domain because $X^2 = 0$ in the latter quotient ring, so $\langle 2,X^2\rangle$ is not a prime ideal.
Upon edit, you should ask yourself if we can write $\langle 5,X^2\rangle = \langle P\rangle$ for some $P \in \mathbb{Z}[X]$. In particular, this means we can find polynomials $Q,R\in \mathbb{Z}[X]$ with $5 = PQ$ and $X^2 = PR$. From $5 = PQ$, we see that both $P$ and $Q$ must be constant polynomials. Say, $P = n$ for some $n \in \mathbb{Z}$. But also $X^2 = PR = nR$ so that $n=1$ and $R=X^2$. It follows that $P =1$, but then $\langle 5,X^2\rangle = \langle 1 \rangle = \mathbb{Z}[X]$. This is a contradiction since $X \notin \langle 5, X^2\rangle.$ Hence, the ideal is not principal.