How many uncountable subsets of power set of integers are there?

Try to first argue this about $\Bbb R$. How many uncountable subsets does $\Bbb R$ have?

Then note that because $\Bbb R$ and $\mathcal P(\Bbb Z)$ have the same cardinality, any bijection witnessing this would map uncountable subsets of $\Bbb R$ to uncountable subsets of $\mathcal P(\Bbb Z)$. So the answer would be the same.

Write $P(\Bbb Z)$ as a disjoint union of sets $A$ and $B$ each of cardinality $c$. There are $2^c$ sets $A\cup X$ with $X\subseteq B$, each of which is uncountable.

Yes, the answer is $2^c$. Here's a "proof:" the number of uncountable subsets of $P(\mathbb{Z})$ is $P(P(\mathbb{Z}))$ minus the number of countable subsets of $P(\mathbb{Z})$. The number of those is $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$, which is $<2^c=\vert P(P(\mathbb{Z}))\vert$.

There are a couple steps there that need to be filled in (counting the number of countable subsets, and showing that "big $-$ small $=$ big"), but these are standard results, and if you haven't seen them before they're good exercises.