The problem statement is as follows
Let $\mathscr{T}$ be the set of all possible topologies on a set $X$. Define the binary relation $\leq$ on $\mathscr{T} \times \mathscr{T}$ by $\tau_{1} \leq \tau_{2} \iff \tau_{1}$ is coarser than $\tau_{2}$. Verify that $\leq$ is a partial ordering of $\mathscr{T}$. Does $\mathscr{T}$ have maximal and minimal elements? If so, what are they?
To show the partial ordering properties was easy. I am stuck on showing the maximal and minimal elements. It seems to me that the minimal element is $\tau_{m} = \{ \emptyset, X \}$ and the maximal element is $\tau_{M} = \mathscr{P}(X) = \{ A : A \subseteq X \}$. The definition for minimal and maximal element is the following
Def. Let $(P, \leq)$ be a partially ordered set and $S \subseteq P$. Then, $m \in S$ is a minimal element of $S$ iff. there is no $s \in S$ such that $s \leq m$ and $m \neq s$. An element $M \in S$ is a maximal element of $S$ iff. there is no $s \in S$ such that $M \leq s$ and $M \neq s$.
You're right about the minimal (indiscrete, your $\tau_m$) topology and maximal (discrete, your $\tau_M$) topology.
The proof that they have this property is easy:
Suppose $\tau \le \tau_m$. Then $\tau \subseteq \tau_m$ by definition. Axiom 1 of a topology says that $\emptyset \in \tau$ and $X \in \tau$ which just says $\tau_m \subseteq \tau$. So two inclusions give us $\tau=\tau_m$.
(I use the reformulation of the definition of a minimal element in a poset $P$: $m$ is minimal iff $\forall p \in P: p \le m \to p=m$. For maximal elements we have a similar reformulation).
So suppose $\tau_M \le \tau$. Then $\tau_M \subseteq \tau$ but $\tau \subseteq \mathscr{P}(X) = \tau_M$ is true by the definition of a topology as a family of subsets of $X$ (axiom 0, in a way) and so again $\tau=\tau_M$ is immediate and $\tau_M$ is maximal.
In this poset there are even a minimum and a maximum, a much stronger notion than a maximal or minimal element. The former are unique, the latter need not be.