Just one thing before the main question: you ask "Does this mean the point $(0,0,30,16,64)$ is a vertex (not necessarily a feasible one)?" In fact, for a point in canonical form, this is guaranteed to be feasible, because all the variables are nonnegative, and that is the only thing we need to check aside from the system of equations being satisfied. I wanted to mention this because it will be important later.
Moving on to the main question:
So why does setting $n$ of $n+m$ vars to zero guarantee I get all vertices? Are there no vertices that have no zero coordinate?
First we should understand what the meaning of a vertex is - or rather, what the point of a vertex is. The reason that a point in the middle of our region isn't worth checking for optimality is that there are nearby feasible points in all directions - and some of those directions will improve the objective function, whatever it is.
In fact, to dismiss a feasible solution from consideration, it is enough to know that it lies halfway between two other feasible solutions. Suppose that $x$, $x+y$, and $x-y$ all satisfy our constraints. Then the value of the objective function at $x$ will be exactly halfway between the values of the objective function at $x-y$ and $x+y$, which means either $x+y$ or $x-y$ will be better than $x$.
(Slightly muddying the issue, all three points could be tied for best, but this turns out not to make a difference.)
So, for which feasible points $x$ can we find a $y$ such that $x+y$ and $x-y$ are also feasible?
Let's illustrate this on an example. Say we're looking at the system $$2x_1 + 3x_2 + x_3 = 30 \\
(2/3) x_1 + 2 x_2 + x_4 = 16 \\
(16/3) x_1 + 4 x_2 + x_5 = 64$$ and we've found the solution $x = (x_1, x_2, x_3, x_4, x_5) = (3, 7, 3, 0, 20)$. How can we find a $y = (y_1, y_2, y_3, y_4, y_5)$ such that both $x+y$ and $x-y$ are feasible?
Well, first, we want $y_4 = 0$. If $y_4 \ne 0$, then either $x_4 + y_4$ or $x_4 - y_4$ will be negative.
Second, we want $y$ to satisfy the associate homogeneous equations: $$2y_1 + 3y_2 + y_3 = 0 \\
(2/3) y_1 + 2 y_2 + y_4 = 0 \\
(16/3) y_1 + 4 y_2 + y_5 = 0$$
Finally, we want $y \ne 0$, or else $x+y$, $x-y$, and $x$ will all be the same point, which doesn't count.
Is this possible? Yes! It's a system of three equations in four unknowns $y_1, y_2, y_3, y_5$. The system is not inconsistent, because a homogeneous system can't be inconsistent. There are more unknowns than equations, so a nonzero solution exists. Scale it down to be very very small, if necessary; a nonzero solution like $y = (0.003,-0.001,-0.003,0,-0.012)$ is ideal.
So our point $x = (3,7,3,0,20)$ is halfway between $x+y = (3.003, 6.999, 2.997, 0, 19.984)$ and $x-y = (2.997, 7.001, 3.003, 0, 20.012)$. Since $y$ is so small compared to $x$, we won't get any negative numbers in $x+y$ or $x-y$, and by construction, $x+y$ and $x-y$ still satisfy the system of equations: they're both feasible!
We can always find such a $y$, provided that the number of nonzero $y$-variables we can pick is larger than the number of equations. The number of nonzero $y$-variables is the same as the number of nonzero $x$-variables. So we conclude that a feasible point worth considering can have at most $m$ nonzero variables, where $m$ is the number of equations. Equivalently, if there are $n+m$ variables, at least $n$ should be set to $0$.
