Prove $f(x)=x^8-24 x^6+144 x^4-288 x^2+144$ is irreducible over $\mathbb{Q}$

Your polynomial $f(x)$ takes prime (or $-$ prime) values at $x = \pm 1, \pm 7, \pm 11, \pm 13, \pm 23, \pm 67, \pm 85, \pm 109, \pm 145, \pm 197, \pm 205, \pm 209, \pm 241, \pm 373, \pm 397, \pm 403, \pm 421$. That's $34$ points. If it factored as $f(x)=g(x) h(x)$, one of $g$ and $h$ must be $\pm 1$ at at least $17$ of these $x$, and either $+1$ at at least $9$ points or $-1$ at at least $9$ points. But a non-constant polynomial that takes the same value at $9$ points must have degree at least $9$, and $f$ has degree only $8$.

You just have to show that $f(x)$ is the minimal polynomial of $\eta=\sqrt{(2+\sqrt{2})(3+\sqrt{3})}$ over $\mathbb{Q}$, i.e. to show that $\eta$ is an algebraic number over $\mathbb{Q}$ with degree $8$. It is pretty straightforward to prove that $\eta^2$ is an algebraic number over $\mathbb{Q}$ with degree $4$, hence we just need to rule out $\eta\in\mathbb{Q}(\sqrt{2},\sqrt{3})$. If that were the case, for any large enough prime $p$ such that both $2$ and $3$ are quadratic residues ($p=24k+1$ is a sufficient condition) we would have that $(2+\sqrt{2})(3+\sqrt{3})$ is a quadratic residue too. That contradicts quadratic reciprocity, and for an explicit counterexample, by considering $p=73$ we get that $21^2\equiv 3\pmod{p}$, $41^2\equiv 2\pmod{p}$ but $(2+41)\cdot(3+21)$ is not a quadratic residue $\!\!\pmod{p}$. There are an infinite number of such counterexamples, hence $\eta\not\in\mathbb{Q}(\sqrt{2},\sqrt{3})$ and $f(x)$ is the minimal polynomial of $\eta$ over $\mathbb{Q}$. In particular, $f(x)$ is irreducible over $\mathbb{Q}$.

It is interesting to point out that "standard tricks" do not work here since $f(x)$ factors over any finite field $\mathbb{F}_p$. This kind of polynomial is known as Hilbert polynomial, if I recall it correctly.