Do $\Bbb H$ and $\Bbb R^2$ have the same uniform structure? The hyperbolic upper half plane $\Bbb H$ and euclidean space $\Bbb R^2$ are not isomorphic as metric spaces, which can be see from the fact that in $\Bbb R^2$ for any point not on a geodesic there exists exactly one geodesic going through that point that does not intersect the original geodsic, whereas in $\Bbb H$ there exist infinitely many such geodesics.
I cannot see any such argument that uses information visible to the uniform structure however.

Are $\Bbb H$ and $\Bbb R^2$ isomorphic as uniform spaces?

 A: As uniform spaces, $\mathbb{R}^2$ and $\mathbb{H}$ are not isomorphic.
One way to prove $\mathbb{R}^2$ and $\mathbb{H}$ are not isomorphic as metric spaces is that the area of a disk grows quadratically with the radius in $\mathbb{R}^2$ but grows exponentially with the radius in $\mathbb{H}$. It is possible to make a similar argument using just the uniform structure.
First we define approximate notions of distance and volume for an arbitrary uniform space. Let $X$ be a uniform space, $U\subset X\times X$ an entourage. For $x$, $y\in X$, define
$$
d_U(x,y):=\inf\big\{n\big|\exists x_0,\ldots,x_n\in X: x_0=x,x_n=y, (x_i,x_{i+1})\in U\,\forall i\big\}\in\mathbb{Z}_{\geq 0}\cup\{+\infty\}.
$$
For $A\subset X$, we define
$$
\mu_U(A):=\sup\big\{n\big|\exists x_1,\ldots,x_n\in A:(x_i,x_j)\not\in U\,\forall i\neq j\big\}\in\mathbb{Z}_{\geq 0}\cup\{+\infty\}.
$$
Now we can prove $\mathbb{R}^2$ and $\mathbb{H}$ are not isomorphic as uniform spaces. For $\epsilon>0$, define an entourage $U_\epsilon:=\{(x,y):d(x,y)<\epsilon\}\subset(\mathbb{R}^2)^2$. We similarly define $V_\epsilon\subset\mathbb{H}^2$. For every $\epsilon>0$ and $x_0\in\mathbb{R}^2$,
$$
\lim_{n\to\infty} \frac{\mu_{U_\epsilon}\big(\{x:d_{U_\epsilon}(x,x_0)<n\}\big)}{n^2} <\infty.
$$
The sets $U_\epsilon$ form a fundamental system of entourages for $\mathbb{R}^2$.
Claim: For every entourage $U\subset V_1$ on $\mathbb{H}$,
$$
\limsup_{n\to\infty} \frac{\mu_{U}\big(\{x:d_{U}(x,x_0)<n\}\big)}{n^2}=\infty.
$$
Proof: The set $\{V_\epsilon\}$ is a fundamental system of entourages, so there must be some $\epsilon$ for which $V_\epsilon\subset U$. Then
$$
\begin{align*}
\limsup_{n\to\infty}\frac{\mu_{U}\big(\{x:d_U(x,x_0)<n\}\big)}{n^2}&\geq\limsup_{n\to\infty}\frac{\mu_{V_1}\big(\{x:d_{V_\epsilon}(x,x_0)<n\}\big)}{n^2}\\
&\geq\limsup_{n\to\infty}\frac{\mu\big(B_{n\epsilon}(x_0)\big)/\mu(B_1(x_0))}{n^2}\\
&=\frac{\epsilon^2}{\mu(B_1(x_0))}\limsup_{m\to\infty}\frac{B_m(x_0)}{m^2}=\infty.
\end{align*}
$$
On the last line, the $\mu$ without subscript denotes ordinary measure on $\mathbb{H}$, and $B_r(p)$ is the ball of radius $r$ centered at $p$. We used the fact that $\mu_{V_1}(A)\geq \mu(A)/\mu(B_1(x_0))$ for every set $A\subset\mathbb{H}$.
