Proof that the $sqrt[k]{z}\, z \in \mathbb N$ counts the amount of numbers less than or equal to z with a $k-$exact power

Question

Empirically, it can shown that $$\mathrm{Floor}[\sqrt[k]{z} ] \,, z \wedge k\in \mathbb N $$ is equal to the amount of numbers which have a $k-$exact root. For example, $\sqrt 36 = 6$ means that there are six numbers whose square root is exact (or 2-exact numbers), namely 36, 25, 16, 9, 4, and 1, with corresponding square roots as 6, 5, 4, 3, 2, 1. For $\sqrt[3] 343 = 7$, with the 3-exact numbers being 343, 216, 125, 64, 27, 8 and 1. And so on.
Proof
Proceeding by cases
Case I: $z=y^k$, that is, $z$ an exact power of the $k-$root function $$\sqrt[k]{y^k}=y$$
Case II: $z \gtrapprox y^k$, that is, $z$ is somehow larger than the closest (from below) $k-$exact power. Let $z=y^k+x\,, x\in \mathbb N$, so $$\sqrt[k]{y^k+x}]= y\sqrt[k]{1+\frac{x}{y^k}}]$$ and $\mathrm{Floor}[y\sqrt[k]{1+\frac{x}{y^k}}] = y$.
Case III: $z \lessapprox y^k$, that is, $z$ is slightly smaller than the nearest $k-$ exact power (from above). Choose $z = y_*^k -x$, so $$\mathrm{Floor}[y_*\sqrt[k]{1-\frac{x}{y_*^k}}] = y_*$$ Notice that $y_* < y$ the "$y$" used for cases I and II.
Q.E.D.
QUESTION
Now then, how come that either $y$ or $y_*$ can actually represent the amount of numbers with $k-$exact root? where is the counting capability of this number? I don't quite completely grasp that capability from the way I wrote this proof. So maybe my proof lacks to be written in a more formal way, or in some other way, or perhaps is plainly wrong, because of the explicit or implicit assumptions that I am making, are not those necessary and/or sufficient to express the counting capability that was shown empirically.
Thanks!