$\mathbb{N}^2$ is countable

Pairing function. $\pi(m,n)=\frac{(m+n)(m+n+1)}{2}+m$ is a bijection $\mathbb{N}^2\to\mathbb{N}$.

Surjectivity. Every nonnegative integer is uniquely $\pi(m,n)$ for some $(m,n)$ (enumerate the diagonals $m+n=k$).

Conclusion. $|\mathbb{N}^2|=\aleph_0$.