quantum mechanics - How does commutation between the hamiltonian and angular momentum operator (squared) imply conservation of Angular momentum?

So we are looking at central potentials in QM;

The lecturer poses the question, when is $\textbf{L}$ conserved? He then considers the commutator of $\hat{H}$ and $\hat{L^2}$.

We have;

$$\hat{H}=-\frac{\hbar^2}{2m}\triangledown^2_R + \frac{\hat{L^2}}{2mr^2}+V(\vec{r})$$

where, $\triangledown^2_R=\frac{1}{r^2}\frac{\partial{}}{\partial{r}}(r^2\frac{\partial{}}{\partial{r}})$

$$[\hat{H}, \hat{L^2}]= -\frac{\hbar^2}{2m}[\triangledown^2_R,\hat{L^2}]+\frac{1}{2m}[\frac{\hat{L^2}}{r^2}, \hat{L^2}]+[V(\vec{r}),\hat{L^2}]$$

The first two terms are zero. The last term is zero if $V(\vec{r})$ is ONLY a function of $r$ not $\theta$ and $\phi$.

Thus having a central potential will ensure the commutator yields 0, but how does this imply angular momentum is conserved?

Strictly how does the commutation above imply conservation, if anything it shows that the two operators commute and share eigenstates.

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I am aware that angular momentum is conserved if there is a central potential as we will have a central force, $F(r)\hat{r}$ hence:

$$\frac{d\hat{L}}{dt}=\frac{d}{dt}(\hat{r}\times\hat{p})=\frac{d\hat{r}}{dt}\times\hat{p} + \hat{r}\times\frac{d\hat{p}}{dt}=\hat{v}\times m\hat{v}+\hat{r}\times F(r)\hat{r}=0$$

Hence $\hat{L}$ is conserved.