I have read somewhere that "$K3$ surfaces admitting finite nonsymplectic group actions are projective". Could someone remind me of a proof?
A sketch of the proof goes as follows. Let $G$ be a finite group of nonsymplectic automorphisms on a $K3$ surface $X$. Since $G$ is nonsymplectic, there exists $g \in G$ such that $g \omega \neq \omega$, where $\omega$ is the holomorphic $2$form on $X$. Then, setting $Y:=X/G$, one has $q(Y)=p_g(Y)=0$, since $q(X)=0$ and by the previous remark the holomorphic $2$form $\omega$ does not descend to the quotient. From this, one proves that either $Y$ is rational (i.e, bimeromorphic to $\mathbb{P}^2$) or the minimal desingularization of $Y$ is an Enriques surface. In any case, $Y$ is a compact algebraic surface, so there exists an ample divisor $H$ on $Y$. Finally, the quotient $\pi \colon X \longrightarrow Y=X/G$ is a finite holomorphic map since $G$ is a finite group. It follows that $\pi^*H$ is an ample divisor on $X$, hence $X$ is projective. Remark 1. As pointed out by Rita in her comment, one must be a bit careful since $Y$ can be singular. I skip the details, which can be found in Frantzen's dissertation. Remark 2. One also proves that $G$ is a symplectic grup of automorphism if and only if the minimal desingularization of $X/G$ is again a $K3$ surface. The reason is that in this case $g \omega=\omega$ for all $g \in G$, so $\omega$ descends to the quotient. 

