Conjugacy and the Class Equation · Superchallenge problem

A nonabelian group $G$ has order $16$. Using the $p$-group theorem, $|Z(G)|$ must be a multiple of $2$, so it could in principle be $2$, $4$, or $8$ (it can't be $16$, or $G$ would be abelian). Which of these three candidate values is actually IMPOSSIBLE?