While this is not mandated in the original question, it is more interesting to ask for which finite $n$ there exists a complete theory with exactly $n$ countable models up to isomorphism.
For $n=1$, we can take the theory of dense linear orders (or any other $\omega$-categorical theory, for that matter).
For $n\ge3$, there is an elegant construction due to Ehrenfeucht. Let $D_k$ ($k\ge1$) be the theory of densely $k$-coloured linear orders: that is, its language consists of a binary predicate $<$, and unary predicates $P_1,\dots,P_k$, and its axioms state that $<$ is a linear order without end-points, each element satisfies exactly one of the predicates $P_1,\dots,P_k$, and for each $i=1,\dots,k$, the set of elements satisfying $P_i$ is dense. Using a similar zig-zag argument as for dense linear orders, it is easy to show that $D_k$ is complete, $\omega$-categorical, and it has elimination of quantifiers.
Now, let $T_{k+2}$ be an extension of $D_k$ in a language with extra constants $\{c_n:n\in\omega\}$, and axioms $\{c_n<c_m:n<m\}\cup\{P_1(c_n):n\in\omega\}$. By quantifier elimination for $D_k$, the theory $T_{k+2}$ is complete. Let $A,B$ be countable models of $T_{k+2}$. We can construct a partial isomorphism by mapping $c_n^A$ to $c_n^B$, each interval $(c_n,c_{n+1})_A$ onto $(c_n,c_{n+1})_B$, and similarly for the unbounded interval $(-\infty,c_0)$: indeed, all these intervals are countable models of $D_k$, hence they are isomorphic. Conversely, any isomorphism of $A$ and $B$ must look like that. Thus, the isomorphism type of $A$ is uniquely determined by the isomorphism type of the “remainder”
$$A^*=\{a\in A:\forall n\in\omega\,c_n^A<a\}.$$
This remainder may be empty, or it may be nonempty with no least element (in which case it is the unique countable model of $D_k$), or it may have a least element $a$: in the last case, $(a,\infty)$ is the unique countable model of $D_k$, and the only choice we have is the colour of $a$. Thus, in total, $T_{k+2}$ has exactly $k+2$ countable models up to isomorphism.
Curiously, a theorem of Vaught states that the remaining case, $n=2$, is in fact impossible: there is no complete theory with exactly two countable models.
An outline of the proof is as follows. If $T$ has only two countable models, it has only countably many complete $n$-types for every $n<\omega$, hence it has an atomic model $A$, and a countable saturated model $S$. Since $T$ is not $\omega$-categorical, it has a nonprincipal complete $n$-type $p(\vec x)$ for some $n$. First, notice that $p$ is realized in $S$ but not in $A$, hence $A\not\simeq S$, hence every countable model of $T$ is isomorphic to $A$ or to $S$. Consider the theory $T'=T+p(\vec c)$, where $\vec c$ are new constants. If $M,N$ are countable models of $T'$, their reducts to the language of $T$ must be isomorphic to $S$, hence they are saturated (this property is preserved by expansion with extra constants). Since $T'$ is complete, this implies $M\simeq N$. Thus, $T'$ is $\omega$-categorical, and as such it has only finitely many complete $n$-types for every $n$. But then the same must hold for $T$, a contradiction.
First, note that $T$ is complete. This can be proved either by quantifier elimination or by Vaught's test using $\aleph_0$-categoricity.
Now suppose you have an isolated maximal type over $M$, say $[\psi(x)]$ (where I'm making explicit your convention of using $x$ as the variable in $1$-types). Since $\psi(x)$ is consistent with $T$ and $T$ is complete, $\exists x\,\psi(x)$ is provable in $T$ and therefore true in $M$. Fix some $a\in M$ such that $\psi(a)$ holds in $M$. Then the type over $M$ realized by $a$ contains $\psi(x)$ and therefore includes the type generated by $[\psi(x)]$. Since the latter is a maximal type, it coincides with the type of $a$.
Best Answer
Hint: if a $\Phi$ does not isolate $p$, then there is a formula $\Psi \in p$, such that $\Phi \land \Psi$ define a proper subset of the set defined by $\Phi$. Repeat this argument for $\Phi \land \Psi$. Can this process go to infinity?