I know this question has been asked before and I have seen different aproaches to it. However, I answered in a way that I don't know if it's correct.
First of all I calculated the order of $[9]_{31}$ in $(\mathbb{Z}/31\mathbb{Z})^*$, which is $15$.
Clearly $[0]_{31}$ isn't a solution to the equation, so let us now work in $(\mathbb{Z}/31\mathbb{Z})^*$. If $[m]_{31}$ is a solution to the equation, then $[m]_{31}^3 = [9]_{31}$. So, in $(\mathbb{Z}/31\mathbb{Z})^*$ we have that $|[m]_{31}^3| = |[9]_{31}| = 15$. Then we can conclude that the (multiplicative) order of $[m]_{31}$ has to satisfy the following equation: $$|[m]_{31}^3| = \frac{\operatorname{lcm}(3,|[m]_{31}|)}{3} \Leftrightarrow \operatorname{lcm}(3,|[m]_{31}|) = 15 \cdot 3 =45$$
Now, the set of the divisors of $45$ is $\{1,3,5,9,15,45 \}$ and we have that $|[m]_{31}| $ belongs to this set. We can immediately exclude $|[m]_{31}| = 1$ and $|[m]_{31}| = 45$. The former because $[1]_{31}$ does not solve the initial equation and the latter because $|(\mathbb{Z}/31\mathbb{Z})^*| = 30 \geq |[m]_{31}|$. We can then see that for all the other possibilities of $|[m]_{31}|$, we have $\operatorname{lcm}(3,|[m]_{31}|) \neq 45$. We conclude that the initial equation has no solutions in $\mathbb{Z}/31 \mathbb{Z}$.
Where I'm having doubts: I believe that I didn't do the passage from $\mathbb{Z}/31 \mathbb{Z}$ to $(\mathbb{Z}/31 \mathbb{Z})^*$ correctly. I feel like I might have done something wrong in the part $[m]_{31}^3 = [9]_{31}$ in $\mathbb{Z}/31 \mathbb{Z}$ implies $|[m]_{31}^3| = |[9]_{31}|$ in $(\mathbb{Z}/31 \mathbb{Z})^*$.
Thank you.
