What's the difference between $\mathbb{A}^n$ and $\mathbb{A}^{n+1}$? Besides the obvious difference in topological dimension.
If you want to distinguish between $\mathbb{R}$ and $\mathbb{R}^2$, take an open set in the plane, remove a point, then it's still connected.  That doesn't work in $\mathbb{R}$.  If you want to distinguish between $\mathbb{R}^3$ and $\mathbb{R}^2$, take an open set in $3$-space, remove a point, then its fundamental group is still trivial.  That doesn't work in $\mathbb{R}^2$.  Similarly, you can use the higher homotopy groups to distinguish $\mathbb{R}^n$ and $\mathbb{R}^{n+1}$.
Is there another subtle topological argument to distinguish Zariski $n$-space and Zariski $n+1$-space?  
 A: If we would like imitate the case of the natural topology on the real affine spaces: I shall work in the scheme setting and I shall use the Čech and Grothendieck cohomologies!

I start from $\mathbb{A}^1_{\mathbb{R}}\equiv\mathbb{A}^1$; let $X_1=\mathbb{A}^1\setminus\{(x_1)\}$, via Rabinowitsch's trick one can prove that $X_1$ is isomorphic to hyperbola $H=V(xy-1)\subset\mathbb{A}^2$, in other words $X_1$ is an affine variety.
I recall the Serre's Criterion for the affiness of a scheme (see [B], theorem 7.7.8).

Let $S$ be a quasi-compact scheme. $S$ is affine if and only if for any quasi-coherent $\mathcal{O}_S$-module $\mathcal{F}$ one has $H^1(S,\mathcal{F})=0$.

In particular, without change the name, $H^1\left(X_1,\mathcal{O}_{X_1}\right)=0$.
Again, I recall that for any scheme $S$ and for any $\mathcal{O}_S$-module $\mathcal{F}$, $H^1(S,\mathcal{F})$ is in bijection with $\check{H}^1(S,\mathcal{F})$; therefore, I start to study $\check{H}^1\left(X_n,\mathcal{O}_{X_n}\right)$ for any $n\geq2$, where $X_n=\mathbb{A}^n_{\mathbb{R}}\setminus\{(x_1,\dots,x_n)\}$.
First all, the $X_n$'s are not affine, for $n\geq2$; because they are not the spectrum of $\mathcal{O}_{X_n}(X_n)$!
Indeed, let $\mathcal{U}=\{D(x_k)\subset\mathbb{A}^n\}_{k\in\{1,\dots,n\}}$ an affine open covering of $X_n$, where $D(x_k)=Spec\mathbb{R}[x_1,\dots,x_n]_{x_k}$; considering the Čech cocomplex:
\begin{equation}
0\to C^0\left(\mathcal{U},\mathcal{O}_{X_n}\right)\stackrel{d_0}{\longrightarrow}C^1\left(\mathcal{U},\mathcal{O}_{X_n}\right)\stackrel{d_1}{\longrightarrow}\dots
\end{equation}
where:


*

*$C^q\left(\mathcal{U},\mathcal{O}_{X_n}\right)=\displaystyle\prod_{1\leq i_1<\dots<i_{q+1}\leq n}\mathcal{O}_{X_n}\left(D\left(x_{i_1}\dots x_{i_{q+1}}\right)\right)$,

*$\left(d^qf\right)_{i_1\dots i_{q+2}}=\displaystyle\sum_{j=1}^{q+2}(-1)^{j+1}f_{i_1\dots\widehat{i_j}\dots i_{q+2}\displaystyle|D\left(x_{i_1}\dots x_{i_{q+2}}\right)}$.
Trivially:
\begin{gather*}
\mathcal{O}_{X_n}(X_n)\cong\check{H}^0\left(\mathcal{U},\mathcal{O}_{X_n}\right)=\ker d^0=\\
=\left\{(f_1,\dots,f_n)\in\mathcal{O}_{X_n}\left(D\left(x_1\right)\right)\times\dots\times\mathcal{O}_{X_n}\left(D\left(x_n\right)\right)\mid\\
\forall h\neq k\in\{1,\dots,n\},\,f_{h\displaystyle|D(x_hx_k)}=f_{k\displaystyle|D(x_hx_k)}\right\}=\bigcap_{i=1}^n\mathcal{O}_{X_n}(D(x_i))=\mathbb{R}[x_1,\dots,x_n].
\end{gather*}
At this point, one can state that $X_1$ is not isomorphic to $X_n$ for $n\geq2$; that is, $\mathbb{A}^1$ can not be isomorphic to $\mathbb{A}^n$ for $n\geq2$.
In consequence to Serre's criterion:
\begin{equation*}
\forall n\geq2,\,H^1\left(X_n,\mathcal{O}_{X_n}\right)\neq0;
\end{equation*}
then what are $H^1\left(X_n,\mathcal{O}_{X_n}\right)$'s? The answer is a little bit complicated...
Then I change strategy: by construction $\forall n\geq1,q\geq n,\,C^q\left(\mathcal{U},\mathcal{O}_{X_n}\right)=0$ and therefore $\forall n\geq1,q\geq n,\,\check{H}^q\left(\mathcal{U},\mathcal{O}_{X_n}\right)=0$; in particular:
\begin{equation*}
\forall n\geq1,\,\check{H}^{n-1}\left(\mathcal{U},\mathcal{O}_{X_n}\right)=\ker d^{n-1}_{\displaystyle/\operatorname{Im}d^{n-2}}=\mathcal{O}_{X_n}\left(D(x_1\dots x_n)\right)_{\displaystyle/\operatorname{Im}d^{n-2}}.
\end{equation*}
I recall a Leray's theorem (see [B], theorem 7.7.5)

Let $\mathcal{U}$ be an open covering of a scheme $X$ and $\mathcal{F}$ be an $\mathcal{O}_X$-module. Assume $H^q(U,\mathcal{F})=0$ for any $q\geq1$ and for any finite intersection $U$ of sets in $\mathcal{U}$; then the canonical map $\check{H}^q(\mathcal{U},\mathcal{F})\to\check{H}^q(X,\mathcal{F})$ is bijective for all $q\geq0$.

and the Vanishig Cohomology Theorem for the Affine Schemes (see [B], proposition 7.6.4 and corollary 7.7.7)

Let $S$ be an affine scheme over a ring $R$. Then for any quasi-coherent $\mathcal{O}_S$-module $\mathcal{F}$, one has $\forall q\geq1,\,H^q(S,\mathcal{F})=0$.

By the previous theorem, $\mathcal{U}$ satisfies the hypothesis of Leray's theorem and then
\begin{equation*}
\forall n\geq1,\,\check{H}^{n-1}\left(X_n,\mathcal{O}_{X_n}\right)\cong\check{H}^{n-1}\left(\mathcal{U},\mathcal{O}_{X_n}\right)=\mathcal{O}_{X_n}\left(D\left(x_1\dots x_n\right)\right)_{\displaystyle/\operatorname{Im}d^{n-2}}
\end{equation*}
where $\mathcal{O}_{X_n}\left(D\left(x_1\dots x_n\right)\right)=\mathbb{R}[x_1,\dots,x_n]_{x_1\dots x_n}$.
By construction:
\begin{gather*}
C^{n-2}\left(\mathcal{U},\mathcal{O}_{X_n}\right)\ni\left(\frac{f_1}{x_2^{\alpha_{1,2}}\dots x_n^{\alpha_{1,n}}},\frac{f_2}{x_1^{\alpha_{2,1}}x_3^{\alpha_{2,3}}\dots x_n^{\alpha_{2,n}}},\dots,\frac{f_n}{x_1^{\alpha_{n,1}}\dots x_{n-1}^{\alpha_{n,n-1}}}\right)\mapsto\sum_{j=1}^n(-1)^{j+1}\frac{f_jx_j}{x_1^{\alpha_{j,1}}\dots x_j\dots x_n^{\alpha_{j,n}}}\in C^{n-1}\left(\mathcal{U},\mathcal{O}_{X_n}\right);
\end{gather*}
in other words:
\begin{equation*}
a,b\in\mathbb{R}[x_1,\dots,x_n]_{x_1\dots x_n},\,[a]=[b]\in\check{H}^{n-1}\left(\mathcal{U},\mathcal{O}_{X_n}\right)\iff a-b=\sum_{\underline{i}\in(\mathbb{N}_0)^n\setminus\{\underline{0}^n\}}\frac{r_{\underline{i}}}{x^{\underline{i}}}
\end{equation*}
where:


*

*$r_{\underline{i}}\in\mathbb{R}$,

*for any $\underline{i}=(i_1,\dots,i_n)\in(\mathbb{N}_0)^n,\,x^{\underline{i}}=x_1^{i_1}\dots x_n^{i_n}$;
then:
\begin{equation*}
\forall n\geq2,\,\check{H}^{n-1}\left(X_n,\mathcal{O}_{X_n}\right)=\bigoplus_{\underline{i}\in(\mathbb{N}_0)^n\setminus\{\underline{0}^n\}}x^{-\underline{i}}\mathbb{R}.
\end{equation*}
Let $n>m\geq2$, then $X_n$ is not isomorphic to $X_m$ because, by the previous reasoning, $\check{H}^{n-1}\left(X_n,\mathcal{O}_{X_n}\right)\neq\check{H}^{n-1}\left(X_m,\mathcal{O}_{X_m}\right)=0$; in consequence $\mathbb{A}^m$ can not be isomorphic to $\mathbb{A}^n$.

Bibliography
[B] Bosch S. - Algebraic Geometry and Commutative Algebra (2013), Springer Verlag.
A: I leave my previous cohomolgical proof, and I post an elementary proof!, without Krull dimension theory. ;)

Considering the map
\begin{equation*}
n\geq2,\,\varphi_n:t\in\mathbb{A}^1_{\mathbb{K}}\to\left(t^n,t^{n+1},\dots,t^{2n-1}\right)\in\mathbb{A}^n_{\mathbb{K}};
\end{equation*}
the image $\gamma_n$ of $\varphi_n$ is an algebebraic subset of $\mathbb{A}^n_{\mathbb{K}}$ given by the system of  equations
\begin{equation*}
\begin{cases}
x_1^{n+k-1}x_k-x_2^{n+k-1}=0\\
k\in\{1,\dots,n\}\setminus\{2\}
\end{cases};
\end{equation*}
because $\varphi_n$ is continuous and surjective onto the image, $\gamma_n$ is an irreducible set of (Krull) dimension $1$, that is, it is an irreducible algebraic curve.
By definition:
\begin{equation*}
\mathbb{K}[\gamma_n]=\mathbb{K}[x_1,\dots,x_n]_{\displaystyle/\left(x_1^{n+k-1}x_k-x_2^{n+k-1}\mid k\in\{1,\dots,n\}\setminus\{2\}\right)},
\end{equation*}
let $\mathfrak{m}$ be the maximal ideal of $O=(0,\dots,0)$ in $\mathbb{K}[x_1,\dots,x_n]$ and let $\overline{\mathfrak{m}}$ be the maximal ideal of $O$ in $\mathbb{K}[\gamma_n]$; let $\pi:\mathbb{K}[x_1\dots,x_n]\to\mathbb{K}[\gamma_n]$ the canonical projection, passing to the localization at $\overline{\mathfrak{m}}$, $\varpi=\pi_{\overline{\mathfrak{m}}}$ (the localization of $\pi$ at $\overline{\mathfrak{m}}$) is a surjective morphism of local rings, in other words $\varpi(\mathfrak{m})=\overline{\mathfrak{m}}$!
Easily, one has that
\begin{equation*}
\forall k\in\{1,\dots,n\},\,\overline{x_k}\notin\left(\overline{x_1},\dots,\overline{x_{k-1}},\overline{x_{k+1}},\dots,\overline{x_n}\right)
\end{equation*}
therefore: $\overline{\mathfrak{m}}$ has $n$ generators, $\mathbb{K}[\gamma_n]_{\overline{\mathfrak{m}}}$ is not a regular ring and $O$ is not a regular point of $\gamma_n$!
By definition, the Zariski cotangent space of $\gamma_n$ at $O$ is the $\mathbb{K}$-vector space
\begin{equation*}
\left(T_O\gamma_n\right)^{\vee}=\overline{\mathfrak{m}}_{\displaystyle/\overline{\mathfrak{m}}^2}=\left\{\left[a_1\overline{x_1}+\dots+a_n\overline{x_n}\right]\subset\mathbb{K}[\gamma_n]\mid a_1,\dots,a_n\in\mathbb{K}\right\}\cong\mathbb{K}^n;
\end{equation*}
in a similar way, one proves that $\left(T_O\mathbb{A}^n_{\mathbb{K}}\right)^{\vee}\cong\mathbb{K}^n$.
In this way, let $i$ be the inclusion of $\gamma_n$ in $\mathbb{A}^n_{\mathbb{K}}$, one can consider the $\mathbb{K}$-linear surjective morphism:
\begin{equation*}
\left(d_Oi\right)^{\vee}:[a]\in\left(T_{i(O)}\mathbb{A}^n_{\mathbb{K}}\right)^{\vee}\cong\mathbb{K}^n\to[\varpi(a)]\in\left(T_O\gamma_n\right)^{\vee}\cong\mathbb{K}^n.
\end{equation*}
If $1\leq m<n$ and $\mathbb{A}^m_{\mathbb{K}}$ is regular isomorphic to $\mathbb{A}^n_{\mathbb{K}}$, then $\gamma_n$, up to regular isomorphisms, is an irreducible, singular curve of $\mathbb{A}^m_{\mathbb{K}}$; let $j$ be the inclusion of $\gamma_n$ in $\mathbb{A}^m_{\mathbb{K}}$, by the same reasoning, $\left(d_Oj\right)^{\vee}$ induces a $\mathbb{K}$-linear surjective morphism from $\mathbb{K}^m$ onto $\mathbb{K}^n$: this is a contraddition! Q.E.D. $\Box$
