Let $f(x)=x^2+bx+c$ Let the roots of $f(x)=0$ be $r(\cos\theta \pm i\sin\theta)$ [where $i=\sqrt{-1}$ and $r\cos\theta \not=0$] It will suffice to prove that $[r(\cos\theta + i\sin\theta)]^{m}\not\in\mathbb{Z}$ for any $m\in\mathbb{N}$ Now if $(\cos\theta + i\sin\theta)^{m}=(\cos m\theta + i\sin m\theta) \in\mathbb{Z}$ then $\sin m\theta = 0 \implies \cos m\theta=\pm1$ In that case the roots of $f(x)=0$ are $\pm r$ [where $r\in\mathbb{Z}$] So $c=-r^2$. But $f(0)=c$ is a square-free integer and is not $1$ or $-1$ So we arrive at a contradiction. Hence $f(x^n)$ is irreducible over $ \mathbb{Z}[x] $ [ proven ]

Carolstar9 wrote: It will suffice to prove that $[r(\cos\theta + i\sin\theta)]^{m}\not\in\mathbb{Z}$ for any $m\in\mathbb{N}$ ] Can u please explain the above ? Thank you.

can somebody tell me how I can prove that a polynomial is irreducible over $ Z $?

there is no unique way for proving irreducibility over $\mathbb Z[x]$, but there are some useful lemmas. see the book Polynomials by V.Prasolov for more information.