As $\text{lcm}(m,n) = 2024 = 8 \cdot 253$ and $8 = 2^3$, at least one of the numbers $m$ and $n$ is divisible by $8$. Since $8 \mid 96 = m-n$, the other one must also be divisible by $8$. Both $m$ and $n$ are divisors of $2024$. All divisors of $2024$ that are divisible by $8$ are $8$, $88$, $184$ and $2024$. The only two of these with difference $96$ are $184$ and $88$. A straightforward check shows that $\text{lcm}(184,88) = \text{lcm}(8 \cdot 23, 8 \cdot 11) = 8 \cdot 23 \cdot 11 = 2024$ indeed.

As $\text{lcm}(m,n) = 2024$, both $m$ and $n$ are divisors of $2024$. All divisors of $2024$ are
1, 2, 4, 8, 11, 22, 23, 44, 46, 88, 92, 184, 253, 506, 1012, 2024.
As $m-n=96$, we must have $m > 96$. This observation cuts out all cases except $m = 184, m = 253, m = 506, m = 1012$ and $m = 2024$. As differences between $253$, $506$, $1012$ and $2024$ are larger than $96$, it suffices to check the options $m = 184$ and $m = 253$ which imply $n = 88$ and $n = 157$, respectively. As $157 \nmid 2024$, the only solution can be $(m,n) = (184, 88)$. An easy check shows that $\text{lcm}(184,88) = \text{lcm}(8 \cdot 23, 8 \cdot 11) = 8 \cdot 23 \cdot 11 = 2024$ indeed.