Closed points are dense in $\operatorname{Spec} A$

You're right that a maximal ideal of $A_f$ corresponds to an ideal in $A$ not containing any power of $f$, but this ideal need not be maximal in $A$.

A distinguished open set $D(f)$ is nonempty if and only if $f$ is not nilpotent (since the nilradical is the intersection of all prime ideals). Showing that the closed points are dense amounts to showing that if $f$ is not nilpotent, then there is some maximal ideal which does not contain it. Turning this around, the closed points are dense if and only if the Jacobson radical (the intersection of all maximal ideals) equals the nilradical (the intersection of all prime ideals). This is not true in general, e.g. take any DVR.