Can every infinite set be divided into pairwise disjoint subsets of size $n\in\mathbb{N}$?

The answer is yes, under the axiom of choice, such a partition is possible. There are several ways of seeing this. For example, choice gives us that any set is in bijection with an infinite ordinal. But, for any infinite ordinal $\alpha$, there is a bijection between $\alpha$ and $\alpha\times \{1,\dots,n\}$. The bijection is in fact canonical, in the sense that there is a "uniform" procedure that applied to any infinite ordinal $\alpha$ produces such a bijection. If you are somewhat familiar with ordinal numbers, a proof can be found in this blog post of mine.

Of course, if $\alpha$ is in bijection with $X$, then $\alpha\times \{1,\dots,n\}$ is in bijection with $X\times \{1,\dots,n\}$, so the latter is in bijection with $X$. The required partition is then the image under the bijection of the sets $A_a=\{(a,i)\mid 1\le i\le n\}$, for $a\in X$.

(To mention but one other standard argument, using choice, we have that $X$ and $X\times X$ are in bijection so, invoking the Bernstein-Schroeder theorem, we conclude that $X$ and $X\times\{1,\dots,n\}$ are in bijection as well.)

On the other hand, the answer is no, without the axiom of choice it is in general not possible to produce such a partition. This is addressed at length in this MO answer, with details in the references linked there.

For infinite sets $S$ and finite sets $A$, we have $|S\times A|=|S|$. If we let $A=\{1,\ldots,n\}$, then the images of sets $\{s\}\times A$, $s\in S$, under a corresponding bijection make a partition as desired.