I know the definition of a plane to be:
$(r-r_0)\cdot n = 0$
where $n$ is the vector perpendicular to the plane, $r$ the vector to a given point and $r_0$ the vectors to the points which constitute the plane.
Now I have also encountered this formulation:
$n^\top \cdot x$ = 0
where the $x$ would be all the points of the plane.
My questions: Are these two definitions equivalent? And is then $n^\top \cdot x + b = 0$ a translation regarding the $z$ axis? Thanks!
$n^\top\cdot x=0$ describes a linear hyperplane, i.e. a hyperplane through the origin. To make it affine, you can indeed write $n^\top\cdot x+b=0$, or – more commonly – like this: $n^\top\cdot x=d$ with $d=-b$. If $\lVert n\rVert=1$, i.e. the normal vector has unit length, then $\lvert b\rvert$ would be the distance between origin and the plane, and the sign would indicate the direction. Note that the $z$ axis plays no special role in this, contrary to your assumption.
To see the eqivalence in the affine case, you can transform the equation like this.
$$n^\top\cdot(x-x_0)=n^\top\cdot x+\underbrace{\bigl(-\overbrace{n^\top\cdot x_0}^d\bigr)}_b=0$$
The distinction between $n$ and $n^\top$ is mostly a syntactic one: $n\cdot x$ is the dot product between two vectors, whereas $n^\top\cdot x$ is the matrix product between a row vector and a column vector. The actual result is the same in both cases.
Also note that I assume $x_0$ to be the point defining the plane, and $x$ to be an arbitrary point for which you check whether it is in the plane. So if I take your question literally, my $x_0$ would be your $r$ and my $x$ would be your $r_0$. But my use of the subscript is more common.
There is one other possible approach: in projective geometry, using homogeneous coordinates, $n^\top\cdot x=0$ itself would be enough to describe all hyperplanes. Getting rid of this mixture between addition and multiplication is one of the main benefits of homogeneous coordinates.