The hyperbolic dimension of $f$ is 1 and its maximal hyperbolic set is the unit circle $\mathbb{S}^1$.

First we show that a hyperbolic set for $f$ must be contained in the unit circle $\mathbb{S}^1$. If $E$ is a closed, $f$-invariant set that is not contained in $\mathbb{S}^1$ then it must contain $0$ or $\infty$. The derivative of $f$ vanishing at $0$ and $\infty$, the set $E$ cannot be hyperbolic.

Then it is easily seen that $\mathbb{S}^1$ is hyperbolic, by computing the derivative of $f^n(z) = z^{2^n}$ on $\mathbb{S}^1$. The restriction of the spherical metric to $\mathbb{S}^1$ is equivalent to the constant one, so it enough to observe that $ (f^n)^{'}(z)=2^n z^{2^n - 1}$ and $ \lvert (f^n)^{'}(z) \rvert =2^n $ for $ z \in \mathbb{S}^1 $. The Hausdorff dimension of $\mathbb{S}^1$ is 1.

As any hyperbolic set is contained in $\mathbb{S}^1$, and the Hausdorff dimension is non-decreasing with respect to inclusion, we conclude that the hyperbolic dimension of $f$ is 1.

The hyperbolic dimension of $f$ is 1 and its maximal hyperbolic set is the unit circle $\mathbb{S}^1$.

First we show that a hyperbolic set for $f$ must be contained in the unit circle $\mathbb{S}^1$. If $E$ is a closed, $f$-invariant set that is not contained in $\mathbb{S}^1$ then it must contain $0$ or $\infty$. The derivative of $f$ vanishing at $0$ and $\infty$, the set $E$ cannot be hyperbolic.

Then it is easily seen that $\mathbb{S}^1$ is hyperbolic, by computing the derivative of $f^n(z) = z^{2^n}$ on $\mathbb{S}^1$. The restriction of the spherical metric to $\mathbb{S}^1$ is equivalent to the constant one, so it enough to observe that $ (f^n)^{'}(z)=2^n z^{2^n - 1}$ and $ \lvert (f^n)^{'}(z) \rvert =2^n $ for $ z \in \mathbb{S}^1 $. The Hausdorff dimension of $\mathbb{S}^1$ is 1.

As any hyperbolic set is contained in $\mathbb{S}^1$, and the Hausdorff dimension is non-decreasing with respect to inclusion, we conclude that the hyperbolic dimension of $f$ is 1.

Edit: the length element of the spherical metric is $ds^2 = \frac{dz d\bar{z}}{(1 + \lvert z \rvert^2)^2}$. This implies that the norm of the derivative in this metric is $ \lVert f' \rVert_z = \frac{1 + \lvert z \rvert^2}{1 + \lvert f(z) \rvert^2} \lvert f'(z) \rvert$ .

The hyperbolic dimension of $f$ is 1 and its maximal hyperbolic set is the unit circle $\mathbb{S}^1$.

First we show that a hyperbolic set for $f$ must be contained in the unit circle $\mathbb{S}^1$. If $E$ is a closed, $f$-invariant set that is not contained in $\mathbb{S}^1$ then it must contain $0$ or $\infty$. The derivative of $f$ vanishing at $0$ and $\infty$, the set $E$ cannot be hyperbolic.

Then it is easily seen that $\mathbb{S}^1$ is hyperbolic, by computing the derivative of $f^n(z) = z^{2^n}$ on $\mathbb{S}^1$. The Hausdorff dimension of $\mathbb{S}^1$ is 1.

As any hyperbolic set is contained in $\mathbb{S}^1$, and the Hausdorff dimension is non-decreasing with respect to inclusion, we conclude that the hyperbolic dimension of $f$ is 1.

The hyperbolic dimension of $f$ is 1 and its maximal hyperbolic set is the unit circle $\mathbb{S}^1$.

First we show that a hyperbolic set for $f$ must be contained in the unit circle $\mathbb{S}^1$. If $E$ is a closed, $f$-invariant set that is not contained in $\mathbb{S}^1$ then it must contain $0$ or $\infty$. The derivative of $f$ vanishing at $0$ and $\infty$, the set $E$ cannot be hyperbolic.

Then it is easily seen that $\mathbb{S}^1$ is hyperbolic, by computing the derivative of $f^n(z) = z^{2^n}$ on $\mathbb{S}^1$. The restriction of the spherical metric to $\mathbb{S}^1$ is equivalent to the constant one, so it enough to observe that $ (f^n)^{'}(z)=2^n z^{2^n - 1}$ and $ \lvert (f^n)^{'}(z) \rvert =2^n $ for $ z \in \mathbb{S}^1 $. The Hausdorff dimension of $\mathbb{S}^1$ is 1.

As any hyperbolic set is contained in $\mathbb{S}^1$, and the Hausdorff dimension is non-decreasing with respect to inclusion, we conclude that the hyperbolic dimension of $f$ is 1.