Assuming $a>b$, we have $2021|(7^a-7^b)$, which is equivalent to $2021|7^{b}(7^{a-b}-1)$. Since there does not exist $a, b \in \mathbb{N} \ s.t. \ 2021| 7^{b}$, then $b=0$. Then we have $2021|7^a-1$. This means that $7^a-1\equiv 0 \pmod{2021}$. Since $2021$ and $7$ are relatively prime, by Euler's theorem, we have $a \equiv 2020 \pmod{2021}$.