I have shown it is irreducible.
I've tried considering $\alpha = \bar{x} \in \mathbb{Z}_2[x]$ s.t. $\alpha^4+\alpha^3+1 = 0$.
From my understanding you use the fact that $\alpha^{16-1} = 1$ and hence show that $\alpha^5 \neq 1$ and $\alpha^3 \neq 1$ (as they are the maximal factors of 15).
Working this through reveals:
$\alpha^4 = \alpha^3 + 1$
$\alpha^5 = \alpha^4 + \alpha = \alpha^3 + \alpha + 1$.
This doesn't appear to help me at all. Have I made a mistake somewhere?
Any advice or guidance would be great, thanks!
It is easy to check that your polynomial has no roots in $ \mathbb{F}_2 $, so if it is reducible, it can only split into irreducible quadratic factors. As every quadratic in $ \mathbb{F}_2[X] $ has a root in $ \mathbb{F}_4 $ (why?) this would mean that our polynomial has a root $ \alpha \neq 0 $ in $ \mathbb{F}_4 $. But then, we have
$$ \alpha^4 + \alpha^3 + 1 = \alpha + 2 = \alpha \neq 0 $$
as $ \alpha^3 = 1 $ and $ 2 = 0 $ in characteristic $ 2 $, so there is no such root, and $ X^4 + X^3 + 1 $ is irreducible in $ \mathbb{F}_2[X] $.
The roots of $ f $ all lie in the field $ \mathbb{F}_2[X]/(f) \cong \mathbb{F}_{16} $, and they satisfy $ \alpha^{15} = 1 $. Therefore, as you've remarked, it suffices to check that $ \alpha^3, \alpha^5 \neq 1 $. We have already checked the first case above, to check the second note that
$$ \gcd(X^5 - 1, X^4 + X^3 + 1) = \gcd(-X^4 - X - 1, X^4 + X^3 + 1) = \gcd(-X^4 - X - 1, X^3 - X) $$
so if $ \alpha^5 = 1 $, we must also have $ \alpha^3 = \alpha $ or $ \alpha^2 = 1 $, which implies that $ \alpha = 1 $; impossible.