Ordered Pairs

On solving the equation, we get mn=144. Also,mn= $2 ^ 4 * 3^2.$ If m and n are natural numbers then (4+1)(2+1)=5 * 3=15 ordered pairs of (m,n) are possible. But it is given to us that m,n are integers.Therefore 2*15=30 pairs are possible.

$\color{#D61F06}{\textbf{Sorry I'm reassessing this approach}}$ $\color{#D61F06}{\textbf{for another varieties and generalizing}}$

$\large144=2^4\times3^2$ $\large n=6 , r=2$ $\large^nP_r=^6P_2=\frac{6!}{4!}=\boxed{\color{#3D99F6}{30}\space pairs}$

Or actually to be more accurate it should be nCr multiplied by 2

But in this case only $2×nCr=nPr$

using the fact that $\textbf{r=2}$ so $\textbf{r!=2!}$ will be cancelled by $\textbf{2}$ and we can consider as if second pair was the -ve one

$\large144=2^4\times3^2$ $\large n=6 , r=2$ $\large2\times ^nC_r=2\times ^6C_2=\frac{2\times6!}{4!\times2!}=\boxed{\color{#3D99F6}{30}\space pairs}$