Claim: $P(\mathbb{Z}^+)$ is uncountable.
Proof.
Suppose its not; then we can let $X_1,X_2,...$ be an enumeration of $P(\mathbb{Z}^+)$. Let $g:P(\mathbb{Z}^+)\setminus\{\mathbb{Z}^+\}\to\mathbb{Z}^+$ be a function that maps each proper subset $X\subset\mathbb{Z}^+$ to some $x\in\mathbb{Z}^+$ such that $x\notin X$. Then $\bigcup_{i=1}\{g(X_i)\}\subseteq\mathbb{Z}^+$, so $\bigcup_{i=1}\{g(X_i)\}=X_j$ for some $j$. But then $g(X_j)\in\bigcup_{i=1}\{g(X_i)\}$, so $g(X_j)\in X_j$; but by def of $g$, $g(X_j)\notin X_j$; contradiction. Therefore $P(\mathbb{Z}^+)$ is countable.
Is it valid? Someone told me my method of using $g$ is not the best, but they didn't tell if the proof is valid.