I cannot draw a reasonable picture, so some visualization and drawing is left to you.
Assume that the cone is "upside down," that is, point down, with point at $(0,0,0)$.
Take a horizontal slice through the cone, at height $z$. The cross-section of the cone at height $z$ has a certain radius, which we can call $r(z)$. Then the volume of the cone is $\int_0^h \pi(r(z))^2\,dz.$ As you saw, we want to find a formula for $r(z)$.
Take a vertical slice through the apex of the cone. The cross-section is an isosceles triangle, of height $h$, with the top length equal to $2r$. If we look at the part of the triangle that goes up to height $z$, that part has top length equal to $2r(z)$. The small triangle is similar to the full triangle, and therefore $\frac{2r(z)}{2r}=\frac{z}{h}.$ We conclude that $r(z)=\dfrac{rz}{h}$.
Our volume is therefore $\int_0^h \pi \frac{r^2}{h^2}z^2\,dz.$
Remark: If you do not want to think of the cone as being point down, let $r(z)$ be the radius at distance $z$ from the bottom, that is, at distance $h-z$ from the top. Basically the same argument as the one above shows that $\frac{2r(z)}{r}=\frac{h-z}{h},$ and therefore our volume is $\int_0^h \pi \frac{r^2}{h^2}(h-z)^2\,dz.$ The integral is just a little bit harder, not much, if we make the substitution $u=h-z$.