Let $d$ denote the greatest common divisor of $m$ and $n$. Then $m = d m_1$ and $n = d n_1$, where $m_1$ and $n_1$ are coprime. The least common multiple of $m$ and $n$ is $d m_1 n_1$. We have
$$101 = d + d m_1 n_1 = d(1 + m_1 n_1).$$
Since $1 + m_1 n_1 \ge 2$ and $101$ is prime, we can only have $d = 1$ and $m_1 n_1 = 100$. The numbers $m = m_1$ and $n = n_1$ are coprime. Hence, the one that is divisible by $2$ must in fact be divisible by $4$. The one that is divisible by $5$ is divisible by $25$. All possible solutions are $(1, 100)$, $(4, 25)$, $(25, 4)$ and $(100, 1)$.