The spaces $\beta \mathbb N$ and $\beta\mathbb N\times \beta\mathbb N$ are not homeomorphic.
To derive a contradiction, assume that $\beta N$ and $\beta N\times \beta N$ are homeomorphic. Since $\beta \mathbb N$ is homeomorphic to $\beta(\mathbb N\times\mathbb N)$, we conclude that there exists a homeomorphism $h:\beta(\mathbb N\times\mathbb N)\to\beta\mathbb N\times \beta\mathbb N$. Taking into account that homeomorphisms preserve isolated points, we would conclude that $f=h|\mathbb N^2$ is a bijection of $\mathbb N\times \mathbb N$ and $h$ is a unique continuous extension of $f$. The bijective map $f^{-1}:\mathbb N^2\to\mathbb N^2$ extends to a homeomorphism $\beta(f^{-1}):\beta(\mathbb N^2)\to\beta(\mathbb N^2)$. Then the composition $H=h\circ\beta(f^{-1}):\beta(\mathbb N\times\mathbb N)\to\beta\mathbb N\times\beta\mathbb N$ is a homeomorphism extending the identity embedding $i:\mathbb N\times\mathbb N\to \mathbb N\times\mathbb N\subset\beta\mathbb N\times\beta\mathbb N$. Since $\mathbb N^2$ is dense in $\beta(\mathbb N\times\mathbb N)$, the map $H$ coincides with the unique continuous extension $\beta i$ of the identity embedding $i:\mathbb N\times\mathbb N\to\beta\mathbb N\times\beta\mathbb N$. To complete the proof, it remains to show that the map $\beta i$ is not injective.
Fix any free ultrafilter $\mathcal U_0$ on $\mathbb N$ and consider two distinct ultrafilters: $\mathcal U=\{\{(x,x):x\in U\}:U\in\mathcal U_0\}$ and $\mathcal V=\{\bigcup_{x\in U}\{x\}\times U_x:U\in\mathcal U_0,\;(U_x)_{x\in U}\in\mathcal U_0^U\}$ on $\mathbb N\times \mathbb N$. It can be shown that $H(\mathcal U)=\beta i(\mathcal U)=\beta i(\mathcal V)=H(\mathcal V)$. So, the map $H$ is not injective and can not be a homeomorphism.