Suppose I have $N\in \mathbb{N}$ identical balls and $M\in \mathbb{N}$ identical baskets.
I throw the balls one by one into the baskets. Each basket has equal probability to catch the ball (no missing). Then my question is what's the distribution of the number of balls a basket finally contains?
Is it related to Poisson distribution?
It is not related to Poisson, it is in fact modelled by the Binomial distribution.
You have probability $1/M$ of a ball going into a specific basket which can be modelled as a "success" and the ball going into any other basket being a "failure".
Therefore with N throws the expected value of balls in any basket will be $N/M$ and the probability of $k$ balls being in a specific basket is:
$p(k) = {N \choose k} (1/M)^k ((N-1)/M)^{\rm\it{n-k}}$