The way you framed the problem, it looks like you're talking about rectilinear motion - i.e. straight up or down. Rewrite it as
$$\frac{dr}{dt} = v, \quad \frac{dv}{dt} = -\frac{GM}{r^2}.$$
Since the motion is straight up or down, you can treat the velocity $v$ as a function of the height $r$, instead of as a function of time $t$ ... at least until you reach a stationary point, if you're not going up at or beyond escape velocity.
For infinitesimal change in time, $dt$, you have
$$dr = v dt, \quad dv = -\frac{GM}{r^2} dt.$$
Therefore
$$\frac{dv}{dr} = \frac{-GM/r^2}{v},$$
or
$$v dv = -\frac{GM}{r^2}dr.$$
or
$$d\left(\frac{v^2}{2}\right) = d\left(\frac{GM}{r}\right).$$
Therefore,
$$v^2 = \frac{2GM}{r} + \text{ constant}.$$
The simplest case is escape velocity, where $v → 0$ as $r → ∞$, because then, the integration constant is $0$, and we could write
$$\frac{dr}{dt} = v = \sqrt{\frac{2GM}{r}}. \tag{Escape Velocity}$$
Then, you have
$$\sqrt{r} dr = \sqrt{2GM} dt,$$
or
$$d\left(\frac{2}{3}r^{3/2}\right) = d\left(\sqrt{2GM}t\right).$$
Thus
$$r^{3/2} = \frac{3}{2} \sqrt{2GM} \left(t - t_0\right),$$
or
$$r^3 = \frac{9}{2}GM \left(t - t_0\right)^2,$$
where the integration constant has, without loss of generality, been written as $t_0$.
Compare this with Kepler's Third Law: for elliptical orbits, period-squared goes as mean orbital radius cubed.