The short answer is that $\alpha>1$ cannot happen. Here is why:

For clarity, I will use $z=x+iy$ as the (complex) domain variable, rather than $x$, and let $D^*= D\setminus\{0\}$ denote the punctured unit disk in the complex plane. The equation $\Delta u = e^{2u}$ is the condition that the metric $g = e^{2u}(\mathrm{d}x^2+\mathrm{d}y^2) = e^{2u}\mathrm{d}z\circ\mathrm{d}\bar z$ have Gauss curvature $K=-1$.

Let $U\subset \mathbb{C}$ be the upper half-plane and consider the universal covering map $\phi:U\to D^*$ defined by $\pi(w) = e^{iw} = z$ for $w\in U$. Then $\phi^*g$ is a metric on $U$ that has curvature $-1$. Since the metric $h = (\mathrm{d}w\circ\mathrm{d}\bar w)/\bigl(\mathrm{Im}(w)\bigr)^2$ on $U$ also has Gauss curvature $-1$ and is complete, it follows that there is a holomorphic map $f:U\to U$ such that $f^*h = \phi^*g$, in other words, under the identification $z = e^{iw} $, we have $$ e^{2u}\,\mathrm{d}z\circ\mathrm{d}\bar z = \frac{|f'(w)|^2}{\bigl(\mathrm{Im}(f(w))\bigr)^2}\,\mathrm{d}w\circ\mathrm{d}\bar w $$ Moreover, because the metric $\phi^*g$ is invariant under the deck transformation $w\mapsto w+2\pi$ for $\phi$, it follows that there exist real constants $a,b,c,d$ satisfying $ad-bc=1$ such that $$ f(w+2\pi) = \frac{a\,f(w)+b}{c\,f(w) + d}. $$ By composing $f$ with an appropriate linear fractional transformation in $\mathrm{PSL}(2,\mathbb{R})$, we can reduce to one of the following possibilities:

1. $f(w+2\pi)= f(w)$. In this case, $f(w) = p(e^{iw})=p(z)$ for some holomorphic map $p:D^*\to U$, and, by removable singularities, $p$ must be holomorphic at $0\in D$. Thus, $f(w) = p(z)$ where $p$ is holomorphic on the entire disk. Thus, $$ e^{2u}\,\mathrm{d}z\circ\mathrm{d}\bar z = \frac{\mathrm{d}(p(z))\circ \mathrm{d}(\overline{p(z)})}{\bigl(\mathrm{Im}(p(z))\bigr)^2} = \frac{\bigl|p'(z)\bigr|^2}{\bigl(\mathrm{Im}(p(z))\bigr)^2}\,\mathrm{d}z\circ\mathrm{d}\bar z . $$ Consequently, $u(z) = k\log|z| + w(z)$ where $k\ge0$ is the order of vanishing of $p'(z)$ at $z = 0$ and $w$ is smooth on $D$. (Thus, $\alpha = - k\le 0$ in this case.)

2. $f(w+2\pi) = (\cos\theta\,f(w) - \sin\theta)/(\sin\theta\,f(w) + \cos\theta)$ for some angle $\theta\in (0,\pi)$. Then, setting $p(w) = \bigl(i-f(w)\bigr)/\bigl(i+f(w)\bigr)\in D$, one finds that $p(w+2\pi) = e^{2i\theta}p(w)$, so $p(w) = e^{i\rho w}q(w)$ where $\rho = \theta/\pi\in (0,1)$ and where $q(w+2\pi) = q(w)$, so $q$ can be written in the form $q(w) = s(e^{iw})= s(z)$. Since $|z|^\rho|s(z)| = |e^{i\rho w}q(w)| = |p(w)| < 1$ for $0<|z|<1$ and since $0<\rho<1$, it follows that $s$ has a removable singularity at $z=0$. Tracing through the change of variables, we find $$ e^{2u}\,\mathrm{d}z\circ\mathrm{d}\bar z = \frac{4\,\mathrm{d}\bigl(z^\rho s(z)\bigr)\circ \mathrm{d}\bigl(\overline{z^\rho s(z)}\bigr)}{\bigl(1-|z|^{2\rho}|s(z)|^2\bigr)^2} = \frac{4|z|^{2\rho-2}\bigl|\rho s(z) + zs'(z)\bigr|^2}{\bigl(1-|z|^{2\rho}|s(z)|^2\bigr)^2}\,\mathrm{d}z\circ\mathrm{d}\bar z. $$ Thus, $u(z) = (k{+}\rho{-}1)\log|z| + w(z)$, where $k\ge0$ is the order of vanishing of $s$ at $z=0$ and $w$ is continuous at $z=0$. Thus, $\alpha = 1-\rho-k<1$ in this case.

3. $f(w+2\pi) = f(w) \pm 2\pi$. (This is two cases, depending on the sign). Then the function $p(w) = e^{if(w)}$ takes values in $D^*$ and satisfies $p(w+2\pi) = p(w)$, so $p(w) = q(e^{iw})$ for some holomorphic $q:D^*\to D^*$, and $q$ must have a removeable singularity at $z=0$. In fact, $q$ must have a zero at $z=0$, otherwise $f$ would have to satisfy $f(w+2\pi) = f(w)$. Now using $e^{if(w)} = q(z)$, we find, since $q(z) = z^ks(z)$ where $k\ge1$ and $s(0)\not=0$, that $$ e^{2u}\,\mathrm{d}z\circ\mathrm{d}\bar z = \frac{\bigl|q'(z)\bigr|^2} {|q(z)|^2\bigl(\log|q(z)|\bigr)^2}\,\mathrm{d}z\circ\mathrm{d}\bar z . $$ Thus, $u(z) = -\log |z| - \log\log|q(z)| + w(z)$, where $w$ is smooth at $z=0$. Thus, $\alpha = 1$ in this case.

4. $f(w+2\pi) = \lambda\,f(w)$ where $\lambda\not=1$ is real and positive. Then, using techniques as above, we now establish that there is a constant $\mu>0$ (that depends on $\lambda$) and a holomorphic mapping $s:D^*\to A_\mu$ where $A_\mu\subset \mathbb{C}$ is the annulus consisting of those $w\in\mathbb{C}$ satisfying $e^{-\mu\pi/2}<|w|^2<e^{\mu\pi/2}$, such that $$ e^{2u}\,\mathrm{d}z\circ\mathrm{d}\bar z = \frac{4\,\sec^2\bigl(\log |s(z)|^{2/\mu}\bigr) |s'(z)|^2}{\mu^2|s(z)|^2} \mathrm{d}z\circ\mathrm{d}\bar z. $$ Again $s$ must have a removeable singularity at $z=0$, and, of course, $s$ does not vanish in $D$ because its image is in $A_\mu$. Thus, $u(z) = k\log|z| + w(z)$ where $w$ is smooth and $k\ge0$ is the order of vanishing of $s'$ at $z=0$. Thus, in this case, $\alpha = -k \le 0 < 1$.

In summary, we have $\alpha \le 1$ in all cases, i.e., $\alpha>1$ does not occur.

The short answer is that $\alpha>1$ cannot happen. Here is why:

For clarity, I will use $z=x+iy$ as the (complex) domain variable, rather than $x$, and let $D^*= D\setminus\{0\}$ denote the punctured unit disk in the complex plane. The equation $\Delta u = e^{2u}$ is the condition that the metric $g = e^{2u}(\mathrm{d}x^2+\mathrm{d}y^2) = e^{2u}\mathrm{d}z\circ\mathrm{d}\bar z$ have Gauss curvature $K=-1$.

Let $U\subset \mathbb{C}$ be the upper half-plane and consider the universal covering map $\phi:U\to D^*$ defined by $\pi(w) = e^{iw} = z$ for $w\in U$. Then $\phi^*g$ is a metric on $U$ that has curvature $-1$. Since the metric $h = (\mathrm{d}w\circ\mathrm{d}\bar w)/\bigl(\mathrm{Im}(w)\bigr)^2$ on $U$ also has Gauss curvature $-1$ and is complete, it follows that there is a holomorphic map $f:U\to U$ such that $f^*h = \phi^*g$, in other words, under the identification $z = e^{iw} $, we have $$ e^{2u}\,\mathrm{d}z\circ\mathrm{d}\bar z = \frac{|f'(w)|^2}{\bigl(\mathrm{Im}(f(w))\bigr)^2}\,\mathrm{d}w\circ\mathrm{d}\bar w $$ Moreover, because the metric $\phi^*g$ is invariant under the deck transformation $w\mapsto w+2\pi$ for $\phi$, it follows that there exist real constants $a,b,c,d$ satisfying $ad-bc=1$ such that $$ f(w+2\pi) = \frac{a\,f(w)+b}{c\,f(w) + d}. $$ By composing $f$ with an appropriate linear fractional transformation in $\mathrm{PSL}(2,\mathbb{R})$, we can reduce to one of the following possibilities:

1. $f(w+2\pi)= f(w)$. In this case, $f(w) = p(e^{iw})=p(z)$ for some holomorphic map $p:D^*\to U$, and, by removable singularities, $p$ must be holomorphic at $0\in D$. Thus, $f(w) = p(z)$ where $p$ is holomorphic on the entire disk. Thus, $$ e^{2u}\,\mathrm{d}z\circ\mathrm{d}\bar z = \frac{\mathrm{d}(p(z))\circ \mathrm{d}(\overline{p(z)})}{\bigl(\mathrm{Im}(p(z))\bigr)^2} = \frac{\bigl|p'(z)\bigr|^2}{\bigl(\mathrm{Im}(p(z))\bigr)^2}\,\mathrm{d}z\circ\mathrm{d}\bar z . $$ Consequently, $u(z) = k\log|z| + w(z)$ where $k\ge0$ is the order of vanishing of $p'(z)$ at $z = 0$ and $w$ is smooth on $D$. (Thus, $\alpha = - k\le 0$ in this case.)

2. $f(w+2\pi) = (\cos\theta\,f(w) - \sin\theta)/(\sin\theta\,f(w) + \cos\theta)$ for some angle $\theta\in (0,\pi)$. Then, setting $p(w) = \bigl(i-f(w)\bigr)/\bigl(i+f(w)\bigr)\in D$, one finds that $p(w+2\pi) = e^{2i\theta}p(w)$, so $p(w) = e^{i\rho w}q(w)$ where $\rho = \theta/\pi\in (0,1)$ and where $q(w+2\pi) = q(w)$, so $q$ can be written in the form $q(w) = s(e^{iw})= s(z)$. Since $|z|^\rho|s(z)| = |e^{i\rho w}q(w)| = |p(w)| < 1$ for $0<|z|<1$ and since $0<\rho<1$, it follows that $s$ has a removable singularity at $z=0$. Tracing through the change of variables, we find $$ e^{2u}\,\mathrm{d}z\circ\mathrm{d}\bar z = \frac{4\,\mathrm{d}\bigl(z^\rho s(z)\bigr)\circ \mathrm{d}\bigl(\overline{z^\rho s(z)}\bigr)}{\bigl(1-|z|^{2\rho}|s(z)|^2\bigr)^2} = \frac{4|z|^{2\rho-2}\bigl|\rho s(z) + zs'(z)\bigr|^2}{\bigl(1-|z|^{2\rho}|s(z)|^2\bigr)^2}\,\mathrm{d}z\circ\mathrm{d}\bar z. $$ Thus, $u(z) = (k{+}\rho{-}1)\log|z| + w(z)$, where $k\ge0$ is the order of vanishing of $s$ at $z=0$ and $w$ is continuous at $z=0$. Thus, $\alpha = 1-\rho-k<1$ in this case.

3. $f(w+2\pi) = f(w) \pm 2\pi$. (This is two cases, depending on the sign). Then the function $p(w) = e^{if(w)}$ takes values in $D^*$ and satisfies $p(w+2\pi) = p(w)$, so $p(w) = q(e^{iw})$ for some holomorphic $q:D^*\to D^*$, and $q$ must have a removeable singularity at $z=0$. In fact, $q$ must have a zero at $z=0$, otherwise $f$ would have to satisfy $f(w+2\pi) = f(w)$. Now using $e^{if(w)} = q(z)$, we find, since $q(z) = z^ks(z)$ where $k\ge1$ and $s(0)\not=0$, that $$ e^{2u}\,\mathrm{d}z\circ\mathrm{d}\bar z = \frac{\bigl|q'(z)\bigr|^2} {|q(z)|^2\bigl(\log|q(z)|\bigr)^2}\,\mathrm{d}z\circ\mathrm{d}\bar z . $$ Thus, $u(z) = -\log |z| - \log\log|q(z)| + w(z)$, where $w$ is smooth at $z=0$. Thus, $\alpha = 1$ in this case.

4. $f(w+2\pi) = \lambda\,f(w)$ where $\lambda\not=1$ is real and positive. Then, using techniques as above, we now establish that there is a constant $\mu>0$ (that depends on $\lambda$) and a holomorphic mapping $s:D^*\to A_\mu$ where $A_\mu\subset \mathbb{C}$ is the annulus consisting of those $w\in\mathbb{C}$ satisfying $e^{-\mu\pi/2}<|w|^2<e^{\mu\pi/2}$, such that $$ e^{2u}\,\mathrm{d}z\circ\mathrm{d}\bar z = \frac{4\,\sec^2\bigl(\log |s(z)|^{2/\mu}\bigr) |s'(z)|^2}{\mu^2|s(z)|^2} \mathrm{d}z\circ\mathrm{d}\bar z. $$ Again $s$ must have a removeable singularity at $z=0$, and, of course, $s$ does not vanish in $D$ because its image is in $A_\mu$. In fact, we now see that this case cannot happen since $D$ is simply-connected, so $s:D\to A_\mu$ can be lifted back to $U$ under the covering map $U\to A_\mu$, implying that $f(w+2\pi) = f(w)$, which is covered by Case 1.

In summary, we have $\alpha \le 1$ in all cases, i.e., $\alpha>1$ does not occur. (Note, by the way, that this argument shows that the order $\alpha$ is always well-defined for any isolated singularity.)