Easy proof that $\mathfrak c=\lvert P(\mathbb Z)\rvert$...

Given a real number, $\alpha,$ take the set of all integers:

$$I_\alpha=\left\{2^{n+1}\lfloor n\alpha\rfloor +2^n \mid n\in\mathbb N\right\}$$

Show the function $\mathbb R\to\mathbb P(\mathbb Z)$ defined as $\alpha\mapsto I_\alpha$ is one-to-one. This is because every non-zero integer can be written uniquely in the form $2^{n+1}k+2^n$, so $I_{\alpha}$ encodes the pairs $(n,\lfloor n\alpha\rfloor)$, and $\lim_{n\to\infty} \frac{\lfloor n\alpha\rfloor}{n}=\alpha$, so if $I_\alpha=I_\beta$, then $\alpha=\beta.$

The reverse direction: Given $I\subseteq \mathbb N$, define:

$$\alpha_{I}=\sum_{n\in I} 10^{-n}$$

Prove that $I\mapsto \alpha_I$ is a one-to-one function $P(\mathbb N)\to\mathbb R.$

Then use a simple argument to prove that $|P(\mathbb N)|=|P(\mathbb Z)|.$

Here is a way to show an equivalent, but different statement. SHow that the power of positive integers has the same cardinality as the set of real numbers in the interval $[0,1]$.

Represent each real number in this interval in base 2: a typical element will look something like $0.100100011011101\ldots$ Consider the set of position numbers after the decimal point for every 1 in the above expansion. For the above example it goes like $\{1,4,8,9,11,12,13,15,\ldots\}$. Clearly given a subset of positive integers one can consider the corresponding infinite binary string having 1's at the appropriate slots as specified by this subset.

Of course there are minor issues coming from the fact that the expansion for a number is not unique. But this should not be a problem with some modification