What are solutions (if any) of the following system for $n_i,m_i \in \mathbb{Z}_{\ge 0}$ and $i=1,2$ ?
$$\left\{ \begin{array}{ll} m_1n_2 + m_2n_1 = n_1n_2 \\ n_i^2 \ge 1+m_jn_j \text{ for } i \neq j \\ n_i \text{ divides } 1+m_jn_j \text{ for } i \neq j \end{array} \right.$$
There are none. Well, except some trivial stuff.
Third condition essentially means that $n_1$ and $n_2$ are coprime (all right, it means more than that, but the rest we won't need).
Second condition is not needed at all.
First condition says:
$$m_1n_2 + \underbrace{m_2n_1 = n_1n_2}_{\text{divisible by }n_1}$$
which means $m_1n_2$ is also divisible by $n_1$, which means $m_1$ alone is, which in turn means that $m_1\geqslant n_1$ (unless $m_1=0$, which we'll discuss later). By similar logic, $m_2\geqslant n_2$, and so the LHS of the first equation hopelessly exceeds the RHS.
Now what if $m_1=0$? From the first equation it follows that $m_2=n_2$, and the question boils down to: $n_1|n_2^2+1,\; n_2|1$, which gives the only (up to permutation) solution: $n_1=2,\;m_1=0,\;n_2=m_2=1$.