Let $X$ be a non-empty set, $\Omega$- is some signature. $\Omega$ - terms are all elements from $X$, $0$-arity symbols from $\Omega$, and $\forall \omega\in\Omega$ of arity $n$ and $a_1,\ldots,a_n\in X$ - are terms, then $\omega(a_1,\ldots,a_n)$ is also a term. $F(\Omega,X)$ - is the set of all terms over $X$.
Task: Prove that for every $x\in X$ the set $F(\Omega,X)\setminus \{x\}$ is a maximal subalgebra of the free algebra $F(\Omega,X)$ and that $F(\Omega,X)$ has no other maximal subalgebras.
What I tried: Let $F(\Omega,X)\setminus \{x\}$ is not a maximal subalgebra of the algebra $F(\Omega,X)$, then Exists $A$ that is is a maximal subalgebra and $A \neq F(\Omega,X)\setminus \{x\}$ so it has to be сlosure under operations in $\Omega$. $\forall a,b\in X,\omega\in\Omega: \omega(a,b)\in A\Leftrightarrow A=F(\Omega,X)$. So $F(\Omega,X)\setminus \{x\}$ is a maximal subalgebra of the algebra $F(\Omega,X)$. But I don't know if it's correct and what to do with "$F(\Omega,X)$ has no other maximal subalgebras"
Let $\mathbf F$ denote the term algebra $F(\Omega,X)$.
It is clear that $\mathbf F\setminus\{x\}$ is a sub-algebra of $\mathbf F$, for each $x \in X$, and thus a maximal one (because it's a maximal proper subset).
Now, let $A$ be a proper sub-algebra of $\mathbf F$ and $t(x_1,\dots,x_n) \in \mathbf F \setminus A$.
Then, for some $1 \leq i \leq n$, we have $x_i \notin A$, whence $A \leq \mathbf F \setminus \{x_i\}$.
So $A$ is a maximal sub-algebra of $\mathbf F$ iff the only such term $t(x_1,\ldots,x_n)$ takes the form $x_i$, for some $i$, and then $A$ has the form that you describe.