Finite-field equality conjecture for nilpotency degrees

Let ${\mathbb F}$ be a finite field and ${\mathbb L}$ an infinite field such that

$$\operatorname{char}({\mathbb F})=\operatorname{char}({\mathbb L}).$$

Let $C_{n,d}^{\mathbb F}$ and $C_{n,d}^{\mathbb L}$ denote the nilpotency degrees of the relatively free associative algebras with identity $x^n=0$ over these fields. Finite-field equality conjecture. The nilpotency degrees are equal:

$$C_{n,d}^{\mathbb F}=C_{n,d}^{\mathbb L}.$$

The source says that this equality holds for $n=2,3$ and all $d$, and for $n=4,5$ when $d=2$. No general proof or refutation is supplied.