 Mahler-measure norm argument at measure two

Let an irreducible $f\in \mathbb{Z}[X]$ have $M(f)=2$ and a root outside the closed unit disk. Its leading coefficient must have absolute value one. The product $B$ of all roots outside the unit disk is then an algebraic integer with $|B|=2$; conjugate pairs make $B$ real, so $B=\pm 2$. The quotient of the norm of any root by $B$ is both rational and an algebraic integer, and its absolute value is at most one. It is therefore $\pm 1$, so every root has absolute norm two.