Twisting the Egyptian Fractions!

$\Rightarrow \frac{1}{m}+\frac{4}{n}=\frac{1}{12}$

$\Rightarrow 12n+48m=mn$

$\Rightarrow mn-48m-12n+\color{#3D99F6}{576}=\color{#3D99F6}{576}$

$\Rightarrow (m-12)(n-48)=\color{#3D99F6}{576}$

Now we just need to find pair of factors of $\color{#3D99F6}{576}$ having one of the factor as an odd number ,

$\Rightarrow \color{#3D99F6}{576}=(576,1),(192,3)$ and $(64,9$ )

So now the pairs $(m,n)$ are $\Rightarrow (588,49),(204,51)$ and $(76,57)$ are only possible pairs.

Answer : $\huge \boxed{3}$ pairs.

$\large{\frac1x + \frac1y = \frac{1}{12}} \\ \Rightarrow \large{\frac{n + 4\cdot m}{n\cdot m} = \frac{1}{12}} \\ \Rightarrow \large{\frac{n\cdot m}{n + 4\cdot m} = 12} \\ \Rightarrow n\cdot m = 12\cdot n + 48\cdot m \\ \Rightarrow n\cdot m - 12\cdot n = 48\cdot m \\ \Rightarrow n\cdot (m - 12) = 48\cdot (m - 12 + 12) \\ \Rightarrow n\cdot (m - 12) = 48\cdot (m - 12) + 48\cdot 12 \\ \Rightarrow n\cdot (m - 12) - 48\cdot (m - 12) = 576 \\ \Rightarrow (m - 12)\cdot (n - 48) = 2^6\cdot 3^2$

Since $n$ is an odd number, then $(n - 48)$ is an odd number too.

Hence $(n - 48) \in \Big\{3^0, 3^1, 3^2\Big\} \Rightarrow n \in \Big\{49, 51, 57\Big\}$ , so the pairs are: $\boxed{(m, n) = \Big\{(588, 49), (204, 51), (76, 57)\Big\}}$