maximality of an ultrafilter 
Let $\mathscr{F}$ be a filter on a set $I$. Prove that $\mathscr{F}$ is an ultrafilter if and only if there does not exist any filter $\mathscr{F^*}$ on $I$ such that $\mathscr{F^*} \supsetneq \mathscr{F}$.
Proof: Let $\mathscr{F}$ be an ultrafilter and suppose there exist a filter $\mathscr{F^*}$ on $I$ such that $\mathscr{F^*} \supsetneq \mathscr{F}$. Then by definition of an ultrafilter, for each $E \subset I$, either $E \in \mathscr{F}$ or $I \setminus E \in \mathscr{F}$. But that would imply that for each $E \subset I$, either $E \in \mathscr{F^*}$ or $I \setminus E \in \mathscr{F^*}$ which implies $\mathscr{F^*}$ is an ultrafilter.

I am stuck after this. I was going to conclude that an ultrafilter cannot contain an ultrafilter but I can't find anything that proves my statement.  Is this case or can an Ultrafilter on a set $I$ contain another ultrafilter on the set $I$?
I am also having a really hard time proving the other direction.
Edit:
Okay I was pointed to some questions similar to this and I am still a little confused. So from what I read I should conclude that this contradicts the maximality of ultrafilters, but I am trying to prove maximality so why can I use this?
I'm also still confused how to start the other direction?
 A: Suppose that $E\in\mathscr{F}^*\setminus\mathscr{F}$. Then in particular $E\notin\mathscr{F}$, so $I\setminus E\in\mathscr{F}$. And $\mathscr{F}\subseteq\mathscr{F}^*$, so $I\setminus E\in\mathscr{F}^*$. But $E\cap(I\setminus E)=\varnothing$, so $\varnothing\in\mathscr{F}^*$, which is impossible.
A: Your quoted "proof" is no proof at all. It even starts out wrong.
Suppose $\mathscr{F}$ is an ultrafilter, i.e.
$$\forall E \subseteq I: E \in \mathscr{F} \text{ or } I\setminus E \in \mathscr{F}\tag{1}$$
We want to prove that there is no filter $\mathscr{F}^\ast$ such that $\mathscr{F}^\ast \supsetneq \mathscr{F}$. so suppose $\mathscr{F}^\ast$ would be such a filter. As we have a proper inclusion of families of subsets, there must be some $ E \subseteq I$ such that $E \in \mathscr{F}^\ast$ and $E \notin \mathscr{F}$. That last fact, combined with $(1)$, which we know for $\mathscr{F}$, we conclude that $I \setminus E \in \mathscr{F}$ (the other option is explicitly ruled out after all).
As $\mathscr{F} \subseteq \mathscr{F}^\ast$ we also have $I\setminus E \in \mathscr{F}^\ast$. But then $\emptyset = E\cap (I\setminus E) \in \mathscr{F}^\ast$ as filters are closed under intersections, but this is a contradiction as $\emptyset$ is not allowed to be a member of a filter.
So the assumption of the proper inclusion leads to a contradiction, so $\mathscr{F}^\ast$ cannot exist. QED
