 Solution

Use the reduction criterion: a primitive integer-coefficient polynomial whose leading coefficient remains nonzero modulo a prime number $p$ is irreducible over $\mathbb{Q}$ if its reduction is irreducible over ${\mathbb{F}}_p$. This follows from Gauss lemma for polynomials, since a nontrivial factorization over the integers would reduce to a nontrivial factorization of the same degrees. For a monic polynomial the factors may be taken monic, so neither degree can drop.

Modulo $2$ the polynomial is $\overline{f}=t^4+t+1$. Its values at $0$ and $1$ are both $1$, so it has no linear polynomial factor. A reducible quartic polynomial without a linear polynomial factor must factor into two irreducible quadratic polynomials. The only monic irreducible quadratic polynomial over ${\mathbb{F}}_2$ is $q=t^2+t+1$: among monic quadratic polynomials a nonzero constant is necessary, and $t^2+1$ has a root, while $q$ has none.

But $t^2\equiv t+1(\mathrm{mod}q)$, so $t^3\equiv 1$ and $t^4\equiv t$. ConsequentlyThus $q$ does not divide $\overline{f}$, excluding the remaining possible factorization. We obtain