There's actually an incredible way of proving it that is quite elementary.

First, we note that $\arctan(2)$ can be represented in the complex number $1 + 2i$ as $\sqrt{5} e^{i \arctan(2)}$

The question of proving whether $\arctan(2)$ is a rational multiple of $\pi$ is now framed as checking if the complex number $(1 + 2i)^n$ is ever real for all $n \in \mathbb{N}$. If it is real, then the argument of $\sqrt{5} e^{i \cdot n *\arctan(2)}$ must be a multiple of pi. In essence, if $(1 + 2i)^n$ is ever real, then $\arctan(2) \cdot n = m \cdot \pi \implies \arctan(2) = \frac{m}{n} \pi$.

Why was it useful to phrase the problem in this manner? Let's try and define a recurrence relation for the terms of $(1 + 2i)^n$. Suppose $a_1 = 1$ and $a_2 = 2$, then define $a_n$ to be the real part of $(1 + 2i)^n$ and $b_n$ to be the imaginary part.

$a_{n+1} + ib_{n+1} = (1+2i) \cdot (1 + 2i)^n = (1+2i) \cdot (a_n + ib_n) \\
a_{n+1} = a_n - 2b_n \\
b_{n+1} = 2a_n + b_n$

Okay, I don't like our recursion having two variables in it. Let's try and mess with it a bit.

$a_n = a_{n-1} - 2b_{n-1} \\
b_n = 2a_{n-1} + b_{n-1}$

Substituting the first equation into the $b_{n+1}$ equation, we get
$b_{n+1} = 2a_{n-1} - 4b_{n-1} + b_n$

Now we want to get rid of $a_{n-1}$, so let's substitute the second equation after solving for $a_{n-1}$
$b_{n+1} = 2b_n - 5b_{n-1}$

Nice, we got it all in one variable.
It may look tempting to solve the recurrence relation, but we can do better.
Let's take both sides mod 5.

$b_{n+1} \equiv 2b_n \pmod 5$

Why did we do this? Well, we know the first term, $b_1$, is 2. Therefore, every term after that will just be a power of 2 (mod 5).

$b_n \equiv 2^n \pmod 5$

If there exists some term $b_k = 0$, then it will be 0 (mod 5), because 5 divides 0. (If $b_k = 0$, then that means there exists a natural number $k$ such that $(1+2i)^k$ is real)

But powers of 2 can never be congruent to 0 (mod 5). Therefore, $(1+2i)^n$ is never real and thus $\arctan(2)$ is not a rational multiple of $\pi$.