Answer is $m=2021$ We have $$\sum_{i = 1}^{2021} m^{a_{i}} = n = \sum_{i = 1}^{2021} (m+1)^{b_{i}}$$ If $m=2021;\; a_{i}=1;\; b_{1}=0$ and $b_{j}=1$ for $j>1$ we get $$2021 \cdot 2021 = n = 2020 \cdot 2022 + 1$$which is true. If $m>2021$ then $$n \equiv \sum_{i = 1}^{2021} (m+1)^{b_{i}} \equiv \sum_{i = 1}^{2021} 1 \equiv 2021 (mod \; m)$$ On the other hand $m^k \equiv 0 (mod \; m)$ when $k$ is not zero and $m^k \equiv 1 (mod \; m)$ when $k=0$ Therefore $$n \equiv \sum_{i = 1}^{2021} m^{a_{i}} \equiv 2021$$only if $a_{i}=0$ and therefore $n$ must equal to $2021<m$

We generalise. When the sum of $k\geq 4$ non-negative integer powers, we claim the maximal value is $m=k$. The construction is \[1\cdot n^2 + (n-1)\cdot n^1 = 2n^2-n = 1\cdot (n+1)^2 + (n-4)\cdot (n+1)^1 + 3\cdot (n+1)^0 \] We now show this is maximal. To do this, take mod $k$ of the expression \[\sum_{i=1}^k m^{x_i} = \sum_{i=1}^k (m+1)^{y_i}\Longrightarrow (\# x_i = 0)\equiv k \pmod{m}\]Firstly, note that $(\# x_i = 0)\leq k$. And since we AFTSOC that $m>k$ exists, we cannot have $(\# x_i = 0)=k-m$. Thus, $(\# x_i = 0)=m$, which forces $n=k$, so $m<n$ and we're done. $\blacksquare$.

We claim that the answer is $m=2021.$ This realizes for $2021^2=n=2020\cdot 2022 +1.$ Assume that $m>2021.$ Let $k+\sum_{i=1}^{2021-k}m^{\alpha_i}=n=\sum_{i=1}^{2021}m^{\beta_i}$ for some $k\in \mathbb{Z}^+,$ $\alpha_i>0,\beta_i \geq 0.$ Since $k\equiv 2021 \pmod m$ we obtain $k=2021\implies n=2021<m,$ contradiction. This finishes our proof.