$\aleph_0$-categoricity implies each finite equivalence class is bounded by some natural number In a homework assignment, I am asked to prove the following:

Let $R$ be an equivalence relation on $\mathbb{N}$ such that $Th((\mathbb{N}, R))$ is $\aleph_0$-categorical.
Prove: there exists $n$ such that no finite equivalence class of $(\mathbb{N},R)$ has more than $n$ elements.

I tried proving by contradiction. We suppose that for every $n$, there is a finite equivalence class with $n$ or more elements. To arrive at the conclusion that $Th((\mathbb{N}, R))$ is not $\aleph_0$-categorical, I was thinking about showing that there exists a (un)countable amount of 1-types. However, I have difficulties understanding what 1-types are. Even with this intuitive explanation, I still think they are hard to understand.
Nonetheless, I made an attempt. As far as I understand, 1-types can be intuitively seen as sets of formulas with one free variable that distinguishes elements. Thus I came up with the idea of making formulas stating:

"There exists an equivalence class with exactly $n$ elements".

But I doubt that having countably many of such formulas with one free variable will give me countably many 1-types. Can someone help me with this problem? I thank you in advance.
 A: The statement "there exists an equivalence class with exactly $n$ elements" does not have any free variables. It is close though, to something that will work: we can express "$x$ has exactly $n$ many elements that are in the same equivalence class". You can even write this down in a single formula $\varphi_n(x)$.
So how do we get infinitely many types? It should be clear that for $n \neq k$ the formulas $\varphi_n(x)$ and $\varphi_k(x)$ are contradictory, we will use this to produce infinitely many different types. If there are arbitrarily large finite equivalence classes then there are distinct $(n_i)_{i \in \mathbb{N}}$ such that for each $i \in \mathbb{N}$ there is an equivalence class of size $n_i$. For each $i \in \mathbb{N}$ we let $a_i$ be such that $a_i$ is in an equivalence class of size $n_i$. Write $p_i(x) = \operatorname{tp}(a_i)$ for the ($1$-)type of $a_i$. By construction we have $\varphi_{n_i}(x) \in p_i(x)$. So by the above discussion we have that $p_i(x) \neq p_j(x)$ for all $i \neq j$ in $\mathbb{N}$ and we have found our infinitely many distinct $1$-types.
