Vtyjkyuk

I would like some help with developping a better intuitive understanding of why $mathbbN^2$ is countable while $2^mathbbN$ is not. As far as I understand we can write a 1:1 enumerating function for both.

I am able to follow Cantor's theorem proof. But I don't know how to directly compare the power set result to that of $mathbbN^2$. The proof is shown in multiple other questions but I cannot find any answer or textbook that compares $mathbbN^2$ with $2^mathbbN$ the way I would like. I understand that for infinite countable $mathbbN$, the Cantor pair function becomes surjective while a 1:1 enumeration of the power set does not. It may be injective but not surjective. Why does it not become surjective the same way as the Cantor pair as we move from a finite subset of $mathbbN$ to the countable infinite superset?

For $mathbbN^2$, I can use the Cantor pair function (a bijection) to enumerate all possible pairs. For a finite set the function is not surjective but for an infinite set it is. Now, for $2^mathbbA$, we could build a 1:1 function, also not surjective for finite sets, as follows (consider $A=c,b,a$):
$$
beginarrayc c
0 mapsto & 000\
hline
1 mapsto a & 001\
2 mapsto b & 010\
3 mapsto b,a & 011 \
hline
4 mapsto c & 100\
5 mapsto c,a & 101 \
6 mapsto c,b & 110 \
7 mapsto c,b,a & 111
endarray
$$
We simply convert the enumerating integer to binary and each zero or one indicates presence of the element in that position (of $A$).
In the above, we would quickly need huge numbers ($2^n$) but so what?The Cantor pair function is also not surjective (we need $n^2$) for a finite set.

I am probably overlooking something.. please point it out to me

EDIT:
Thanks to the answer and comments below, there is actually a simple example I see now: with Cantor's function we prove surjectivity going from $mathbbN^2 rightarrow mathbbN$ by showing we can find some pair $in mathbbN^2$ to pruduce any $n in mathbbN$. But in the above power set scheme, all we have to do is show one example where it does not work. So consider for example the infinite set of all even numbers in $2^mathbbN$. It is a subset of $mathbbN$ so it is in the power set. But we cannot use the above enumeration when the set has infinite length.

My intuition is quite satisfied by the fact that one can put $mathbb N^2$ in a list, by putting the elements in an array, and winding one's way back and forth in a spiral pattern ($2$ variations), starting from the upper left hand corner.

Whereas, with $2^mathbb N$, we have a pretty easy identification of subsets of $mathbb N$ and binary representations of the reals.
Couple this Cantor's diagonal argument, and we are done...