There is a good reason to use $z$ instead of $e^{sT}$. Before starting with any analysis, let me remind you that in analysis of signals and systems we are interested in analyzing the frequency spectrum of the signal, i.e. the Laplace transform on the imaginary line $s = jw$. And since your signal $x[n]$ is discrete, then its frequency spectrum its periodic, so its more general define $s=jwT$.
Now, let $X(z) = \mathcal{Z}\{x[n]\}$ of a causal or non-causal discrete signal $x[n]$, i.e.
$$ X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n}. $$
Since $z\in\mathbb{C}$ we have $z = |z| e^{j\arg z}$. Without loss of generality we rewrite $|z| = r$ and $\arg z = wT$, i.e. $z=r e^{jwT}$ (note that not necessarily $r=1$). Then
$$ \begin{aligned}
X(z) &= \sum_{n=-\infty}^{\infty} x[n] z^{-n}\\
&= \sum_{n=-\infty}^{\infty} x[n] (r e^{jwT})^{-n}\\
% &= \sum_{n=-\infty}^{\infty} (x[n] r^{-n}) e^{-njwT}\\
&= \sum_{n=-\infty}^{\infty} (x[n] r^{-n}) (e^{jwT})^{-n},
\end{aligned} $$
which implies $X(z) = \left. \mathcal{L}\{x[n]r^{-n}\} \right\rvert_{s=jwT} = \mathcal{F}\{x[n]r^{-n}\}$. As a consequence, $X(z)$ is a Fourier transform more generic than the Fourier transform $X(e^{jwT}) = \mathcal{F}\{x[n]\}$ of our signal of interest.
So, if the convergence radius of $X(z)$ is less than unity then $X(e^{jwT})$ does not exist and therefore its Fourier transform does not either, which represents a problem because there are many signals with this problem of convergence, e.g. non-causal signals such as a digital image filter. Therefore, it is convenient (and even necessary in non-causal signals) to use the Z-transform.
Or informally, use $z$ instead of $\left. e^{sT} \right\rvert_{s=jwT} = e^{jwT}$ whenever you can.
We also recommend to see this link about radius convergence.
At this point, it is clear that the Z-transform has the same objective as the Laplace transform: ensure the convergence of the transform in some region of $\mathbb{C}$, where the Z-transform does it for discrete signals and Laplace transform for continuous signals.
Best Answer
Due to the identity
$\frac{d}{dt}e^{-st}=-se^{-st}$
analogous to $Ax=\lambda x$ for the Laplace transform you have the eigenvector $e^{-st}$ with ist Eigenvalue $-s$. In functional Analysis, functions are regarded as generalized vectors and Operators like Differentiation as generalized matrices.
For example, you have the differential equation: $\frac{d}{dt}x + x = (\frac{d}{dt}+1)x = 0$. For some functions $x(t)$ you will see that the Operator $\Delta:=\frac{d}{dt}+1$ Change the function (similar as a Matrix makes a vector completely different). But when you do the Laplace transform to this differential equation you will get $sL(x)+L(x)=(s+1)L(x)$ and you observe that the Laplace transform diagonalized the Operator $\Delta$, because the new "Basis" $L(x)$ is only multiplied by $s+1$ (similar to Eigenvalues in linear algebra). The Operator $\Delta$ was diagonalized. Hence, the Basis transform was performed via Laplace transform $x \rightarrow L(x)$.