I'm a bit confused because I was previously under the impression, from a course on signals and systems, that the Z transform was the discretized version of the Laplace transform. However, I just read this post about the relationship between the Z transform and the Laplace transform, which, if I'm understanding it correctly, essentially defines $z$ as $e^{-sT}$ where $T$ is the sampling time, and the Z transform as $L(z)$ (or possibly as $L(\sum z)$; I'm a bit unclear on that part). But if that's the case,
How does the Z transform model discrete systems any better than the regular Laplace transform does? If, as it seems, $z$ is just a parameterized variable of a function of s, I don't see how it's useful specifically for modeling discrete functions of time, since $s$ is continuous.
How does this relate to the $x(n)$ function discussed in the above linked question?
The post states
This is precisely the definition of the unilateral Z-transform of the discrete function $x[n] \ $. $$ X(z) = \sum_{n=0}^{\infty} x[n] z^{-n} $$ with the substitution of $z \leftarrow e^{s T} \ $.
But if $x(n)$ is discrete and the Z transform is just a parameterized version of the Laplace transform, which is itself an integral transform, how does it even make sense to apply it to discrete functions? That is, how does it make sense to integrate a discrete function, given that integration is typically defined on continuous functions?