The highschool Math textbooks that I have seen, derive the equation of a 2D line
y = mx + c
or, equivalently,
ax + by + c = 0
via a Cartesian proof that starts out by defining the slope of a 2D line.
However, when deriving the equation of a 3D-plane, the books use Vector Algebra-based proof to first come up with a vector equation, from which the Cartesian equation is derived in a subsequent step.
I would like to see a pure, Cartesian-only proof for the equation of a plane, if one exists. Does it?
A related question would be: If ax + by + c = 0 is the equation of a 2D-line, then why isn't ax + by + cz + d = 0 the equation of a 3D-line but instead happens to be the equation of a 3D plane?
That's because in a line in the $xy$ plane you're defining $y$ in terms of $x$ and therefore you will need a point and the slope (or two points that will provide the slope in turn).
However, when it comes to 3D space and planes, you're plugging two values, $x$ and $y$, to get a value $z$. Again you will need a point and the slope but now it makes no sense talking about the slope since there are two variables to deal with. Instead, you'll have partial slopes that control the how the plane tilts for each axis. These partial slopes (partial derivatives) can be though as a vector and that's why the proof you're talking about involves vector algebra.
The reason you don't get a line in 3D is precisely that: you have two dependent variables, that is, to get your $z$ you have to plug values from the whole $xy$ plane. To get the line you have to constraint the possible values for both $x$ and $y$ and that could be done by intersecting two planes like in this answer:
What is the equation for a 3D line?