$7^m-p\cdot2^n|7^m+p\cdot 2^n\Leftrightarrow 7^m-p\cdot2^n|2\cdot 7^m\overset {(7^m-p\cdot2^n,2)=1}{\Rightarrow}7^m-p\cdot2^n|7^m $ So $7^m-p\cdot 2^n=7^k,0\leq k<m$ Case 1:$k=0$ ,so $7^m-p\cdot 2^n=1\Rightarrow 1-p\cdot2^n\equiv 1\pmod 3\Leftrightarrow 3|p\cdot2^n\Leftrightarrow p=3$, Therefore $7^m-3\cdot 2^n=1$ If $n=1$ then $7=7^m\Leftrightarrow m=1$ and $\dfrac{7^m + p \cdot 2^n}{7^m - p \cdot 2^n}=\dfrac{7+3\cdot 2}{1}=13=prime$ If $n\geq 2$ then $7^m\equiv 1\pmod 4\Leftrightarrow 2|m$.Let $m=2f$ $(7^f-1)(7^f+1)=3\cdot2^n$.Since $(7^f-1,7^f+1)=2$ there are 4 cases : $(1): 7^f-1=2,7^f+1=3\cdot 2^{n-1}\Rightarrow 7^f=3 $,which is impossible $(2): 7^f-1=6,7^f+1=2^{n-1}\Rightarrow 8=2^{n-1}\Leftrightarrow n=4,f=1$ $(3): 7^f-1=2^{n-1},7^f+1=6\Rightarrow 7^f=5$,which is impossible $(4): 7^f-1=3\cdot 2^{n-1},7^f+1=2$,which is impossible So $n=4,m=2f=2$ and $\dfrac{7^m + p \cdot 2^n}{7^m - p \cdot 2^n}=\dfrac{49+3\cdot 16}{1}=97=prime$ Case 2:$k\neq 0\Rightarrow 7|p\Leftrightarrow p=7$ So $7^m-7\cdot 2^n=7^k\Rightarrow k=1$ : $7^m-7\cdot 2^n=7\Leftrightarrow 7^{m-1}-2^n=1$,which has no solutions ($\pmod 4..\Rightarrow 2|m-1$ etc) So $\dfrac{7^m + p \cdot 2^n}{7^m - p \cdot 2^n}\in \mathbb{Z}\Leftrightarrow \dfrac{7^m + p \cdot 2^n}{7^m - p \cdot 2^n}=13,97=primes$