It is similar to http://www.artofproblemsolving.com/community/q1h1279751p6727211 But what was problem in contest ?

If $p=2$ $2^n>28 \to n>4$ $2|m \to 8|2^n-m^3=7p^2$ - contradiction, so $p \neq 2$ $m^3=0,\pm 1 mod 7, 2^n=1,2,4 mod 7$ ,so $2^n=1 mod 7\ to n=3k$ $m^3+7p^2=2^{3k}$ $7p^2=(2^k-m)(2^{2k}+m2^k+m^2)$ $2^k-m<2^k<2^{2k}+m2^k+m^2$ , and $p^2>7$ so Case 1: $2^k-m=7, 2^{2k}+m2^k+m^2=p^2$ $2^k=m+7$ $m^2+14m+49+m^2+7m+m^2=p^2$ $3m^2+21m+49=p^2\geq 73 \to p>7$ $p^2=1,4,2 mod 7, 3m^2=0,3,5,6 mod 7$ ,so $3m^2-p^2 \neq 0 mod 7$ - contradiction. Case 2: $2^k-m=p, 2^{2k}+m2^k+m^2=7p$ $2^k=m+p$ $m^2+2mp+p^2+m^2+pm+m^2=7p$ $3m^2+3pm+p^2=7p\geq p^2+3p+3$ $p^2-4p+3 \leq 0 \to p=3$ $3m^2+9m+9=21 \to m=1$ $2^k=m+p=4 \to k=2,n=6$ $1^3+7*3^2=2^6$ $(1,6,3)$ - solution.

firstly we must check by modulo 7.then we got n divisible 3 then irreplace $7p^2=2^n-m^3$. then easily we can prove if p>5 the equation has no solution

RagvaloD wrote: It is similar to http://www.artofproblemsolving.com/community/q1h1279751p6727211 But what was problem in contest ? Yes but this is the true form. Other way we coludn't solve equation.

By considering $\pmod{7},$ we get $3\mid n$.Obviously $n\ge 3,$then $n=3k(k\in \mathbb N)$Thus $7p^2=(2^k-m)(2^{2k}+2^km+m^2)$ $(2^k-m)^2<2^{2k}+2^km+m^2=(2^k-m)^2+3\cdot 2^km$ Let $A=2^k-m,B=2^{2k}+2^km+m^2,$then $(A,B)=(1,7p^2),(7,p^2),(p,7p),(p^2,7)$ Case1:$(1,7p^2)$ $7p^2=1+3\cdot 2^km.$Since $p$ is odd,then $ord_2(7p^2-1)=1\implies k=1\implies n=3.$But since $m^3+7p^2>7\cdot 2^2=28>8=2^3,$this is absurd.Thus there is no solution. Case2:$(7,p^2)$ $p^2=49+3\cdot 2^km=49+3m(m+7)\implies p^2\equiv 3m^2\pmod{7}$ Since $49<p^2\implies p>7,$then $p\neq 0\pmod{7}$.Thus $p^2\equiv 1,2,4\pmod{7}.$ $3m^2\equiv 0,3,5,6\pmod{7}.$Then $p^2\equiv 3m^2\pmod{7}$ is absurd.Thus there is no solution. Case3:$(p,7p)$ $p^2<7p\implies p<7$ $7p=p^2+3\cdot 2^km\implies 2^km=\frac{7p-p^2}{3}$ Case3-1:$p=2$ $2^km=\frac{10}{3}$ which is absurd.Thus there is no solution. Case3-2:$p=3$ $2^km=4\implies (k,m)=(1,2),(2,1).$Only $(k,m)=(2,1)$ satisfies the condition.Then $(m,n,p)=(1,6,3)$ which satisfies the condition. Case3-3:$p=5$ $2^km=\frac{10}{3}$ which is absurd.Thus there is no solution. Case4:$(p^2,7)$ $p^4<7$ which is absurd.Thus there is no solution. From case1,2,3,4,the answer is $\boxed{(m,n,p)=(1,6,3)}\blacksquare$