Let $x \in Z$ be a singular point and let $i_x$ be the Cartier index of $K_Z$ at $x$. Note that $i_x$ divides $I$.

Since $Z$ has quotient singularities, there exists a finite group $G \subset \mathrm{GL}_2(\mathbb{C})$ and an analytic neighborhood $x \in U \subset Z$ which is isomorphic to a neighborhood of the origin in $\mathbb{A}^2/G$. Let $G_0 \subset G$ be the kernel of the determinant map $\operatorname{det} : G \to \mathbb{G}_m$ and $H = G/G_0$ its image (which is necessarily a cyclic group). An element $g \in G$ acts on $dx \wedge dy$ via $\operatorname{det}(g)$ so the local index $i_x$ of $K_Z$ is the exponent of $H$ is equal to the order of $H$. The canonical cover of $\mathbb{A}^2/G$ is $\mathbb{A}^2/G_0$ with its natural $H$ action and the map $\mathbb{A}^2/G_0 \to \mathbb{A}^2/G$ is the quotient by $H$.

This gives a local model for $\bar{Z} \to Z$ around each point of $\pi^{-1}(x)$. The ramification index is $\#H = i_x$ so for degree reasons the size of $\pi^{-1}(x)$ is $I/i_x$. The groups $G_0$ are subgroups of $SL_2(\mathbb{C})$ so $\bar{Z}$ has ADE singularities and the minimal resolution $\mu:Z' \to \bar{Z}$ is crepant. On the other hand, by construction $K_\bar{Z} \sim 0$ so $Z'$ is a smooth minimal surface with $K_{Z'} \sim 0$. Note also that quotient singularities are rational so the irregularity $q(\bar{Z})= q(Z')$.

By the classification of complex minimal surfaces, $Z'$ is either an abelian surface or a K3 surface. If $Z'$ is abelian then $Z' \to \bar{Z}$ must be an isomorphism since abelian surfaces contain no rational curves. To get $q(Z) = 0$, the cyclic group of order $I$ must act on $\bar{Z}$ in such a way so that the two dimensional representation on $H^1(\bar{Z}, \mathcal{O}_\bar{Z})$ has no invariants. Moreover, the locus with nontrivial stabilizers is isolated, and the stabilizer at any point has no elements with trivial determinant. I'm not sure off the top of my head if there exist abelian surfaces with such a cyclic group action.

In the other case, $Z'$ is a K3 surface with a cyclic group action which permutes configurations of ADE curves in orbits of size $I/i_x$ according to the local structure above and $\bar{Z}$ is obtained by contracting these curves to ADE singularities. In this case there is no condition on irregularity since $q(Z')= 0$. In either case the topology of the cover is determined.

Let $x \in Z$ be a singular point and let $i_x$ be the Cartier index of $K_Z$ at $x$. Note that $i_x$ divides $I$.

Since $Z$ has quotient singularities, there exists a finite group $G \subset \mathrm{GL}_2(\mathbb{C})$ and an analytic neighborhood $x \in U \subset Z$ which is isomorphic to a neighborhood of the origin in $\mathbb{A}^2/G$. Let $G_0 \subset G$ be the kernel of the determinant map $\operatorname{det} : G \to \mathbb{G}_m$ and $H = G/G_0$ its image (which is necessarily a cyclic group). An element $g \in G$ acts on $dx \wedge dy$ via $\operatorname{det}(g)$ so the local index $i_x$ of $K_Z$ is the exponent of $H$ is equal to the order of $H$. The canonical cover of $\mathbb{A}^2/G$ is $\mathbb{A}^2/G_0$ with its natural $H$ action and the map $\mathbb{A}^2/G_0 \to \mathbb{A}^2/G$ is the quotient by $H$.

This gives a local model for $\bar{Z} \to Z$ around each point of $\pi^{-1}(x)$. The ramification index is $\#H = i_x$ so for degree reasons the size of $\pi^{-1}(x)$ is $I/i_x$. The groups $G_0$ are subgroups of $SL_2(\mathbb{C})$ so $\bar{Z}$ has ADE singularities and the minimal resolution $\mu:Z' \to \bar{Z}$ is crepant. On the other hand, by construction $K_\bar{Z} \sim 0$ so $Z'$ is a smooth minimal surface with $K_{Z'} \sim 0$. Note also that quotient singularities are rational so the irregularity $q(\bar{Z})= q(Z')$.

By the classification of complex minimal surfaces, $Z'$ is either an abelian surface or a K3 surface. If $Z'$ is abelian then $Z' \to \bar{Z}$ must be an isomorphism since abelian surfaces contain no rational curves. To get $q(Z) = 0$, the cyclic group of order $I$ must act on $\bar{Z}$ in such a way so that the two dimensional representation on $H^1(\bar{Z}, \mathcal{O}_\bar{Z})$ has no invariants. Moreover, the locus with nontrivial stabilizers is isolated, and the stabilizer at any point has no elements with trivial determinant. I'm not sure off the top of my head if there exist abelian surfaces with such a cyclic group action.

In the other case, $Z'$ is a K3 surface with a cyclic group action which permutes configurations of ADE curves in orbits of size $I/i_x$ according to the local structure above and $\bar{Z}$ is obtained by contracting these curves to ADE singularities. In this case there is no condition on irregularity since $q(Z')= 0$. In either case the topology of the cover is determined.

Edit: I just noticed that most of what I wrote above is explained in the linked paper in the remark following the definition of a log Enriques surface.