I claim the answer is $2^{2023}$. Let $d=(m,n)$ where $m=dm_1$ and $n=dn_1$. With this, we have \[ dm_1n_1\mid d^{2022}m_1^{2023} + d^{2022}n_1^{2023} + n_1. \]Note that we have $dn_1\mid d^{2022}m_1^{2023} + n_1$ and in particular $d\mid n_1$ and $n_1\mid d^{2022}$. Clearly $d\ne 1$ as otherwise $n=1$. Take any prime $p\mid d$. Note that $v_p(dn_1) = v_p(d)+v_p(n_1)$. Now the key observation is that for both $2022v_p(d)<v_p(n_1)$ and $2022v_p(d)>v_p(n_1)$, this divisibility is impossible. So, $v_p(n_1)=2022v_p(d)$. Hence, $d\ge 2$ and $n_1\ge 2^{2022}$, so $n\ge 2^{2023}$. For $n=2^{2023}$, we choose $m_1$ such that $m_1\mid n_1(n^{2022}+1)$. This is easy, e.g. $m_1=2^{2022\cdot 2023}+1$ works.