We claim that the answer is $336$. Suppose there exists $n\geq 337$ for which $(n!)^6-6^n$ is divisible by $2022=6\cdot 337$, which is impossible, since $(n!)^6$ is divisible with $337$ and $6^n$ isn't, so $n\leq 336$. If $n=336$: $(n!)^6-6^n$ is clearly divisible with $6$, so we just want to prove that it's also divisible with $337$. By Wilson's Theorem $336!\equiv -1$ (mod $337$), so $(336!)^6\equiv 1$ (mod $337$) and by Fermat $6^{336}\equiv 1$ (mod $337$), so $n=336$ works

lol I'm too slow. Same solution as above. Since the LHS is always divisible by $6$ for $n\ge 3$, we basically just want the largest $n$ satisfying \[ (n!)^6\equiv 6^n\pmod{337}. \]Clearly, $n < 337$. Also, by Wilson's Theorem and Fermat's Little Theorem, \[ (336!)^6 \equiv 6^{336} \equiv 1 \pmod{337}. \]