$a^3-b^3=p \to a-b|p \to a=b+1 \to p=3b^2+3b+1 \equiv 1 \pmod 6$

Suppose that $a,b\in\mathbb N$ and a prime $p$ such that $a^3-b^3=p$. Then \[(a-b)(a^2+ab+b^2)=p.\]We have two cases: Case 1: $a-b=p$ and $a^2+ab+b^2=1$. But this is impossible, cuz we'd have $(a-b)^2+3ab=1$ or $p^2=1-3ab<1$. Case 2: $a-b=1$ and $a^2+ab+b^2=p$ Hence $a=b+1$. Substitute back, \begin{align*} &(b+1)^2+(b+1)b+b^2=p\\ &3b(b+1)+1=p \end{align*}Note that $2\mid b(b+1)$, hence $6\mid 3b(b+1)$. Therefore \[ p=3b(b+1)+1\equiv 1\pmod{6} \]as expected.