Prove that the set $\{ \lfloor n\pi \rfloor, n \in \mathbb{N}\}$ contains an arithmetic progression with $m$ elements for any integer $m\ge 3$.
I did a few terms. The set starts with
$$0, 3, 6, 9, 12, 15, 18, 21, \color{red}{25}, 28, 31,34,37,40,43,\color{red}{47},50,53,\\56,59,62,65,\color{red}{69},72,75,78,81,84,87,\color{red}{91},94,97,100,103,...$$
and so on. For $m\le 8$ the first $8$ elements are an arithmetic progression with common difference $3$.
For $m> 8$, the numbers with red are where the difference with the previous term is $4$ and they happen every 7th term (I guess because $7\pi \approx 22$). But $7 \pi < 22$, so from a certain point on this pattern won't be respected anymore. Does this help in any way?
There's an easy way separate from the path you're looking at. The easy way is to just note that, for every real number $r$ we have $$\inf_{n\in\mathbb N}\{nr\}=0$$ where $\{\cdot\}$ is the fractional part of a number. This fact isn't too hard to prove from scratch, though I won't reproduce a proof here. Then, you can just find some $n$ such that $n\pi = k + \alpha$ where $k\in \mathbb Z$ and $\alpha < 1/m$. Then the floors of $0n\pi,\,1n\pi,\,2n\pi,\ldots,(m-1)n\pi$ are just $0,\,k,\,2k,\ldots,(m-1)k$, which is an arithmetic progression.
Though I'm not sure of a way to use it to solve this particular problem, your observation does connect into deep observations with continued fractions, which can be used to describe the way that the whole sequence $\lfloor n\pi\rfloor$ looks to follow a simple pattern over short periods (e.g. "increases by 3 each step" or "increases by 3 each step, except every seventh step where it increases by 4" and then describing that the "every seventh step" is actually changing between "seven" and "eight" steps somewhat regularly - but the period at which this happens isn't quite constant either). It's a bit much to explain in the compass of an answer, but the patterns you're noticing have some interesting consequences.