Let $p$ be a prime, $\mathbb{F}_p$ the field of $p$ elements. Is it possible to describe $n \times n$ (for $n$ sufficiently large if needed) matrices $M$ over $\mathbb{F}_p$ such that $M^p = 0$? Does the exponent of $p$ have more consequences than just saying that $M$ is nilpotent?
What if we look at a prime power $q$ and matrices over $\mathbb{F}_q$: can we describe matrices $M$ such that $M^p = 0$? Or such that $M^q = 0$?
Edit Any nilpotent matrix $M$ will satisfy $M^n = 0$, as pointed out in the comments. I am hoping for some stronger consequence that apply specifically to matrices that are $p$-nilpotent. Ideally some result of the form "any such matrix is conjugate to something of this form".