I just want to check if the following proof is correct.
Let $f(t)=t^4-32\in \mathbb{Z}[t]$. Then $$f(t)=(t-\sqrt[4]{32})(t+\sqrt[4]{32})(t-i\sqrt[4]{32})(t+i\sqrt[4]{32})$$
but $\sqrt[4]{32}=2\sqrt[4]{2}$. It is clear that $\mathbb{Q}(2\sqrt[4]{2})=\mathbb{Q}(\sqrt[4]{2})$, and $t^4-2\in \mathbb{Q}[t]$ is irreducible by Eisenstein's criterion, thus $$[\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}]=[\mathbb{Q}(2\sqrt[4]{2}):\mathbb{Q}]=4$$ so the minimal polynomial $g(t)\in \mathbb{Q}[t]$ of $2\sqrt[4]{2}$ must also have degree four. Considering $f$ as a polynomial over $\mathbb{Q}$, we must have $g|f$, but both of these have degree four and are monic, therefore $f=g$ and $f$ is irreducible.
This proof looks correct. An alternative method is to appeal to the unique factorisation of polynomials in $\mathbb R [t]$ into products of linear and quadratic factors, so $$f(t) = (t - \sqrt[4]{32})(t + \sqrt[4]{32})(t^2 + \sqrt{32}).$$ If $f$ is reducible in $\mathbb Z [t]$, then it has at least one factor of degree at most $2$ with integer coefficients; however the existence of such a factor would clearly contradict the unique factorisation in $\mathbb R [t]$ above, as none of $t \pm \sqrt[4]{32}$, $t^2 + \sqrt{32}$, or $(t - \sqrt[4]{32})(t + \sqrt[4]{32})$ are in $\mathbb Z [t]$.