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This would contradict the Harris-Mumford(-Eisenbud) theorem that $M_g$ is non-uniruled for $g$ at least $23$. Let $C$ be a general curve of genus $g$. If $C$ is in "Zieve form", then it is the normalization of the (almost certainly) singular curve in $\mathbb{CP}^1 \times \mathbb{CP}^1$, $$D = \{ ([x_0,x_1],[y_0,y_1]) \in \mathbb{CP}^1\times \mathbb{CP}^1 \vert y_0^e f(x_0,x_1) - x_0^dg(y_0,y_1) \}, $$ where $f(x_0,x_1)$, respectively $g(y_0,y_1)$, is a homogeneous polynomial of degree $d$, resp. $e$, such that $f(0,1)$ and $g(0,1)$ are nonzero (or else the defining polynomial factors to a simpler form). By direct computation, the singular points occur where $[x_0,x_1]$ is a multiple root of $f(x_0,x_1)$ and $[y_0,y_1]$ is a multiple root of $g(y_0,y_1)$ or the point is $([0,1],[0,1])$. Moreover, at each point, the local analytic type of the singularity is the same as the plane curve with equation $y^n-x^m$, where $m$, resp. $n$, is the vanishing order of $f(x_0,x_1)$, resp. $g(y_0,y_1)$ at that point. In particular, the "delta invariant" depends only on $(m,n)$. Thus, if you "deform" $f(x_0,x_1)$ and $g(y_0,y_1)$ so that the number and type of multiple roots remains constant, then the normalizations of the corresponding curves in $\mathbb{CP}^1\times \mathbb{CP}^1$ remain of genus $g$. However, the family of such deformations of $(f,g)$ is a rational variety. Precisely, if you write $$ f(x_0,x_1) = (x_1-a_1x_0)^{m_1}(x_1-a_2x_0)^{m_2}\cdots (x_1-a_rx_0)^{m_r}, $$ with $(a_1,\dots,a_r)$ pairwise distinct, then the deformation space for $f$ is just a Zariski open subset of the affine space with coordinates $(a_1,\dots,a_r)$, and similarly for $g(x_0,x_1)$. Since $M_g$ is non-uniruled, this is a contradiction: there is only the constant morphism from a rational variety to $M_g$ whose image contains the general point parameterizing $C$.