Yes, this is straightforward. The Frobenius map $x \mapsto x^2$ generates the Galois group of any finite extension of $\mathbb{F}_2$ (in particular the splitting field of $f(x) = \frac{x^p - 1}{x - 1}$), so $f$ is irreducible if and only if the Frobenius map acts transitively on its roots. Letting $\alpha$ denote one of these roots, it follows that $f$ is irreducible if and only if $p-1$ is the smallest positive integer $k$ such that
$$\alpha^{2^k} = \alpha.$$

Now $\alpha$ by assumption has order $p$ in the multiplicative group of $\overline{ \mathbb{F}_2 }$, so $f$ is irreducible if and only if $p-1$ is the smallest positive integer $k$ such that
$$2^k \equiv 1 \bmod p$$

and this condition is equivalent to $2$ being a primitive root mod $p$.

More generally this argument can be used to work out how a cyclotomic polynomial $\Phi_n(x)$ factors modulo a prime.