For the sake of completeness, I will give an argument, given originally by R. Fokkink (see also here), to show that $\mathbb R^{2n+1}$ is not a perfect square for any positive natural $n$; the argument is given only for $n=1$, i.e., this argument shows that $\mathbb R^3$ is not homeo to $X^2$ for any topological space $X$, but it generalizes nicely, to the general case of $\mathbb R^{2n+1}$ being a perfect square.
By contradiction, assume that there is an actual topological space X such that there is a homeo. between $\mathbb R^3$ and $X^2$.
We now then use the fact that homeomorphisms are either orientation-preserving or orientation-reversing (since $\mathbb R$ is orientable, it makes sense to talk about orientability of X). Associated with the concept of orientability is the concept of the degree of a map, so that, if a homeo. $h$ is orientation-preserving, we assign to $h$ a degree $1$, and if $h$ changes orientation, then we assign a degree of $-1$ to $h$.
This concept of degree satisfies the nice property that, given two homeomorphisms $f,g$, then $\deg(f\circ g)=(\deg f)\times(\deg g)$, so that, in particular, $\deg(h\circ h)=1$, i.e., a composition of a homeo. with itself is orientable.
Now, assume $h$ is the given homeo. between $\mathbb R^3$ and $X^2$, so that we also have a homeo. , say $h_1$ between $X^4\rightarrow \mathbb R^6$ taking coordinates $(a,b,c,d)\rightarrow (e,f,g,h,i,j).$
We will get a contradiction by using the assumption of the existence of a homeo $h$, and producing an automorphism $\mathbb R^6 \rightarrow \mathbb R^6$ , whose square does not preserve orientation, contradicting the properties of deg given above.
Consider this: the linear automorphism $L:X^4\rightarrow X^4$ with: $(a,b,c,d)\rightarrow (d,c,b,a)$, so that the composition $L\circ L$ preserves orientation (and I guess it must somehow be funny too ), with $L \circ L:(a,b,c,d\rightarrow (c,d,a,b)$.
Now, we can pullback this linear automorphism $L\circ L $ of $X^4$ to an automorphism of $\mathbb R^6$ using the homeomorphism $h_1$ above, and we would end up with the auto L\circ L':(e,f,g,h,i,j)\rightarrow (i,j,e,f,g,h): But this last is a linear map, and its determinant is $-1$, so the assumption of the existence of a homeo. $h: \mathbb R^3\rightarrow X^2$we constructed a self-homeomorphism of $\mathbb R^6$ whose square is not orientation-preserving.