Partition of N into infinite number of infinite disjoint sets?

Sure. For example, let $A_n$ be the natural numbers with exactly $n$ ones in their binary expansion.

Alternately, pick your favorite way of decomposing $\mathbb{N}$ into two disjoint infinite subsets $A, B$, and pick a bijection $f : B \to \mathbb{N}$. Then $f(B)$ can be decomposed into two disjoint subsets $A, B$, hence $B$ can be decomposed into two disjoint subsets $f^{-1}(A), f^{-1}(B)$. Rinse and repeat. This argument is fairly general and works for any infinite set which admits a decomposition into two disjoint subsets of the same cardinality as it (which under the Axiom of Choice is all of them).

Maybe we can start slowly, by doing a decomposition of $\mathbb{N}$ into $2$ disjoint sets, say the odds and the evens.

Let's now go for a decomposition into $3$ disjoint sets. Leave the odds alone, and decompose the evens into those divisible by $2$ but no higher power of $2$, and those divisible by $4$. To put it another way, we are using the odds, twice the odds, and the rest.

Continue, and let's introduce some notation. Let $W_0$ be the set of odd positive integers. Let $W_1$ be the set of positive integers which are $2^1$ times an odd number. Let $W_2$ be the set of positive integers which are $2^2$ times an odd number. In general let $W_n$ be the set of integers which are $2^n$ times an odd number.

It is clear that the $W_k$ are all infinite, pairwise disjoint, and that their union is all of $\mathbb{N}$.

Let $$ A_0 = \lbrace 1,3,5,7,9,\ldots \rbrace $$ and $$ A_1 = \lbrace 2^n 1 : n \in \mathbb{N} \rbrace, $$ $$ A_2 = \lbrace 2^n 3 : n \in \mathbb{N} \rbrace, $$ $$ A_3 = \lbrace 2^n 5 : n \in \mathbb{N} \rbrace, $$ $$ A_4 = \lbrace 2^n 7 : n \in \mathbb{N} \rbrace, $$ $$ A_5 = \lbrace 2^n 9 : n \in \mathbb{N} \rbrace, $$ $$ \cdots. $$ Noting that for any two distinct elements $r_1$ and $r_2$ of $A_0$ it holds $2^{n_1}r_1 \neq 2^{n_2}r_2$ $\forall n_1,n_2 \in \mathbb{N}$, we have that the $A_i$ are disjoint. On the other hand, let $2k$, with $k \in \mathbb{N}$, be an arbitrary even natural number. Considering its prime factorization, it is necessarily of the form $2k = 2^n r$, where $n \in \mathbb{N}$ and $r \in A_0$. Hence $2k \in \cup _{i = 1}^\infty A_i$, from which it follows that $\cup _{i = 1}^\infty A_i = \lbrace 2,4,6,8,10,\ldots \rbrace$, and so $\mathbb{N} = \cup _{i = 0}^\infty A_i$, with all the $A_i$ disjoint and countably infinite.

EDIT: Relation to user6312's answer.

The sets $A_i$, $i = 1,2,3,\ldots$, correspond to the rows $$ 2,4,8,16,32, \ldots, $$ $$ 6,12,24,48,96, \ldots, $$ $$ 10,20,40,80,160, \ldots, $$ $$ 14,28,56,112,224,\ldots, $$ $$ 18,36,72,144,288,\ldots, $$ $$ \cdots, $$ while the sets $W_i$, $i=1,2,3,\ldots$, in user6312's answer correspond to the corresponding columns, that is to $$ 2,6,10,14,18,\ldots, $$ $$ 4,12,20,28,36,\ldots, $$ $$ 8,24,40,56,72,\ldots, $$ $$ 16,48,80,112,144,\ldots, $$ $$ 32,96,160,224,288,\ldots, $$ $$ \cdots. $$