Is $\mathbb R^3$ the square of some topological space?

No such space exists. Even better, let's generalize your proof by converting information about path components into homology groups.

For an open inclusion of spaces $X \setminus \{x\} \subset X$ and a field $k$, we have isomorphisms (the relative Kunneth formula) $$ H_n(X \times X, X \times X \setminus \{(x,x)\}; k) \cong \bigoplus_{p+q=n} H_p(X,X \setminus \{x\};k) \otimes_k H_q(X, X \setminus \{x\};k). $$ If the product is $\mathbb{R}^3$, then the left-hand side is $k$ in degree 3 and zero otherwise, so something on the right-hand side must be nontrivial. However, if $H_p(X, X \setminus \{x\};k)$ were nontrivial in degree $n$, then the left-hand side must be nontrivial in degree $2n$.

this blog post refers to some papers with proofs. I've heard Robert Fokkink explain his proof (which is, quoting from this post)

A linear map $\Bbb R^n \to \Bbb R^n$ can be understood to preserve or reverse orientation, depending on whether its determinant is $+1$ or $-1$. This notion of orientation can be generalized to arbitrary homeomorphisms, giving a "degree" $\deg(m)$ for every homeomorphism which is $+1$ if it is orientation-preserving and $-1$ if it is orientation-reversing. The generalization has all the properties that one would hope for. In particular, it coincides with the corresponding notions for linear maps and differentiable maps, and it is multiplicative: $\deg(f \circ g) = \deg(f)\cdot \deg(g)$ for all homeomorphisms $f$ and $g$. In particular (fact 1), if $h$ is any homeomorphism whatever, then $h \circ h$ is an orientation-preserving map.

Now, suppose that $h : X^2 \to \Bbb R^3$ is a homeomorphism. Then $X^4$ is homeomorphic to $\Bbb R^6$, and we can view quadruples $(a,b,c,d)$ of elements of $X$ as equivalent to sextuples $(p,q,r,s,t,u)$ of elements of $\Bbb R$.

Consider the map $s$ on $X^4$ which takes $(a,b,c,d) \to (d,a,b,c)$. Then $s \circ s$ is the map $(a,b,c,d) \to (c,d,a,b)$. By fact 1 above, $s \circ s$ must be an orientation-preserving map. But translated to the putatively homeomorphic space $\Bbb R^6$, the map $(a,b,c,d) \to (c,d,a,b)$ is just the linear map on $\Bbb R^6$ that takes $(p,q,r,s,t,u) \to (s,t,u,p,q,r)$. This map is orientation-reversing, because its determinant is $-1$. This is a contradiction. So $X^4$ must not be homeomorphic to $\Bbb R^6$, and $X^2$ therefore not homeomorphic to $\Bbb R^3$.

and there he also told us the cohomological proof, which generalizes it to all Euclidean spaces of odd dimension.

I didn't know that, but I did know this: we cannot have $S^2 = S\times S$ for any topological space $S$.