Compare both sides $\mod 3$ , we should have $k \equiv n \equiv 1$ ($\mod 2$) (i) $k \geq 3$ $\implies$ $3^n \equiv 7^n \equiv 7 $ ($\mod 8$ ) $\implies$ Contradiction (ii) $k=2$ $\implies$ $m^2 = 1+ 2+ \dots + 2^l = 2^{l+1} -1$ . But $4 \nmid m^2 +1$ so there is no solution in this case . (iii) $k=1$ $\implies$ $9(3^{m-2} - 1) = 7( 7^{n-1} -1)$ If $m>2 , n>1$ then above equation implies $6 \mid m-2$ , $6 \mid n-1$ ($n$ is odd) Now $43 \mid 7^6 -1$ also $7 \mid ord_{43} (3)$ , so $7 \mid m-2$ $\implies$ $7^2 \mid 3^{m-2} - 1$ (by LTE) but It's obviously wrong because $7^2$ can't divide $9(3^{m-2} - 1) = 7( 7^{n-1} -1)$ So the only solution is $(m,n,k,l)=(2,1,1,1)$ .

mahanmath wrote: Compare both sides $\mod 3$ , we should have $k \equiv n \equiv 1$ ($\mod 2$) $7=1(\mod 3)$ $\therefore 7^n\equiv 1(\mod 3) \forall n.$ Then how are you saying $n=1(\mod 2)$. Please explain your argument. Thanks, Faustus.

For $k=1$ we have $3^m-7^n=2$ For $m=1$ no solution For $m=2$ then $n=1$ and so $(2,1,1,1)$ is a solution For $m\ge3$ taking modulo $27$ we have $7^n+2\equiv0(\text{mod} 27)$ as $7^n\equiv7,22,19,25,13,10,16,4,1(\text{mod}27)$ so $n\equiv4(\text{mod} 9)$ taking modulo $37$ we have $7^n\equiv7,12,10,33,9,26,34,16,1(\text{mod} 37)$ and since $n\equiv4(\text{mod} 9)$ we get $3^m\equiv7^n+2\equiv35(\text{mod} 37)$ But $3^m\equiv3,9,27,7,21,26,4,12,36,34,28,10,30,16,11,33,25,1(\text{mod} 37)$ and so there is no solution for $3^m\equiv7^n+2\equiv35(\text{mod} 37)$ which gives no solution for $m\ge3$ For $k\ge2$ taking modulo $4$ gives $m$ and $n$ has the same parity. And by taking modulo $7$ gives $3^m \equiv 2^k (\text{mod} 7)$ checking all possibility gives $m$ an even integer, hence $\exists x,y\in\mathbb{N}$ such that $m=2x$ and $n=2y$ then $(3^x-7^y)(3^x+7^y)=2^k$ and since $gcd(3^x-7^y,3^x+7^y)=2$ then $3^x-7^y=2$ and $3^x+7^y=2^{k-1}$ but the only solution for $(x,y)$ is $(2,1)$ hence $m=4$ , $k=5$ and $n=2$ but then no integer solution for $l$. So no solution for this case The only solution is $(m,n,k,l)=(2,1,1,1)$

mahanmath wrote: Now $43 \mid 7^6 -1$ also $7 \mid ord_{43} (3)$ , so $7 \mid m-2$ $\implies$ $7^2 \mid 3^{m-2} - 1$ (by LTE) but It's obviously wrong because $7^2$ can't divide $9(3^{m-2} - 1) = 7( 7^{n-1} -1)$ Where is the LTE here?

mahanmath wrote: Compare both sides $\mod 3$ , we should have $k \equiv n \equiv 1$ ($\mod 2$) (i) $k \geq 3$ $\implies$ $3^n \equiv 7^n \equiv 7 $ ($\mod 8$ ) $\implies$ Contradiction i don't understand this. plz explain me.

matrix41 wrote: Where is the LTE here? $m-2=6s$ and $7 \mid s$ so by LTE $v_7 ((3^6)^s -1) = v_7 (3^6 -1 ) + v_7 (s) \geq 2$