I have a question about the relationship between Laplace and Z transformations. By definition Laplace transformation maps signals in the continuous time range modeled as functions $f:[0,\infty[ \to \mathbb{C}$ onto the complex-valued frequence space via
$$ f \mapsto \mathcal{L} \{f\}(s) := \int_{0}^{\infty} f(t)e^{-st} dt $$
The (unilateral) Z-transformation on the other hand maps a signal in the discrete time domain described by a sequence $(x[k])_{k \in \mathbb{N}$ ($k=0,1,2...$ forms a sampling sequence) via
$$ x[k] \mapsto X(z) = Z\{x[k]\} := \sum_{k=0}^{\infty} x[k] \cdot z^{-k} $$
Now the philosophy is that Z-Trafo is the "discrete version" of the Laplace transformation. What is not clear to me, however, is the exact reason why in Laplace transformation the argument $s$ is exponated to $e^{-st}$, while for the argument $z$ in Z-transformation that's not the case?
Wouldn't it be more natural for sake of compatibility to define the Z-transformation $X(z)$ via $\sum_{k=0}^{\infty} x[k] \cdot e^{-zk}$ to preserve structure compatibility between continuous and discrete versions simply by mimic the continuous Laplace transformation?
Or is there a deeper reason involved?
I found here a discussion dealing with similar problem but the answers there not answer my question. My question is not if there is a relation between $z$ and $s$ in Laplace and Z-transformations (indeed obviously there is one: that's the substitution $z= e^{sT}$ as stated there), but why this transformation should be imposed there.
If we keep on the philosophy that Z-transformation should be the discrete analoga of the Laplace transformation why not define $X(z)$ via $\sum_{k=0}^{\infty} x[k] \cdot e^{-zk}$ as suggested above. The correspondence between continuous and discrete Fourier transformations are established by mimic literally the structure.
What's going wrong with Laplace und Z-transformations if we try to imitate this the same way?