Why are K3 surfaces minimal?

What Beauville says is that "[c]learly, $K\equiv 0$ implies that they are minimal."

A surface $S$ is minimal if any birational morphism $S\to S'$ to any other surface $S'$ is an isomorphism. Thus, suppose that some K3, say $S,$ is not minimal. This means, by definition, that there exists a birational morphism $S\to S'$ which is not an isomorphism. By Theorem II.11 of Beauville, such a morphism can be factored as a (finite, nonzero) sequence of blowups at a point, and by Proposition II.3(iv), $K_S$ is a nonzero effective linear combination of pullbacks of the exceptional divisors of the successive blowups plus the strict transform of $K_{S'}$. Since $\operatorname{Pic}(S)=\operatorname{Pic}(S')\oplus \mathbb Z^m,$ where $m$ is the number of blowups, this gives $K_S\neq 0$ (i.e., we cannot cancel by linear equivalence) thus the contradiction, hence the K3 must be minimal.