We easily see that the given expression is $\binom{x_n}{m}$. Now we see that all numbers $x_n$ have this form $561655\ldots555$ in base $7$, where after the second $6$, there are $n-1$ $5$'s. Using Lucas's theorem with modulo $7$, we get that $m$ must have at most $4$ digits in base $7$ (because $2022=_7 5616$). To maximize $m$ let it have $4$ digits. First, second and third digit must be $1$, because if they were larger, we get that one of $x_1,x_2$ or $x_3$ will be divisible by $7$ by Lucas's theorem, because $\binom{1}{n}=0$ for $n>1$. Same argument goes with the last digit, which can be at most $5$. So maximum of $m$ is $1115$ in base $7$, that is $404$.