According to the question,

$\begin{array}{r}{2}^{\mathrm{m}}-2^{\mathrm{n}}=112\\ \Rightarrow 2^{\mathrm{n}}\left(2^{\mathrm{m}-\mathrm{n}}-1\right)=2^4\times 7\end{array}$

Compering exponents on both side, we get

$2^{\mathrm{n}}=2^4\text{ and }{2}^{\mathrm{m}-\mathrm{n}}-1=7$

Now, $2^{\mathrm{n}}=2^4$

$\Rightarrow \mathrm{n}=4$

And $2^{\mathrm{m}-\mathrm{n}}=8$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{2}^{\mathrm{m}-\mathrm{n}}=2^3\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{m}-\mathrm{n}=3\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{m}-4=3\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{m}=7\end{array}$