By Fermat's Little Theorem for every $m^6\equiv 1$ or $0$ mod $7$, also $a_1=2019 \equiv 3, a_2=2020 \equiv 4, a_3=2021 \equiv 5$ all mod $7$ . We can calculate in mod $7$ $a_4 \equiv 3, a_5\equiv 4, a_6 \equiv 6, a_7\equiv 3, a_8\equiv 4, a_9\equiv 3, a_{10}\equiv 3, a_{11} \equiv 4, a_{12}\equiv 4, a_{13}\equiv 3, a_{14}\equiv 3$ How $a_8\equiv a_{12}, a_9\equiv a_{13}, a_{10}\equiv a_{14}$ in mod $7$ and $a_k$ depends on the above three we generate a pattern and for none $i$, $a_i\equiv 1$ or $0$ mod $7$ so never is a power of six