my approach : first note the solutions : $(0 , 0 , 1) , (1 , 0 , 3)$ then we assume that $x , y ,z > 2$ and we obtain that x and y must be even. This , in turn implies that z must be 8 mod 14 , which means that z is a multiple of 3. if z is even : then let $z = 2k ,y = 2b$ and make $7^x = (2^a - 13^b)*(2^a + 13^b)$ and this has no integer solutions. similarly work wheb z is odd and obtain that there are no more solutions.

Your solution is wrong there is also one solution

sorry , it could be when z is odd.

If $y=0$: $7^x+1=2^z$: two obvious solutions : $(0,0,1),(1,0,3)$ And I don`t know are there other solutions Let $y>0$ $2^z\equiv (-1)^y$(mod 7) has solution only for $z=3z_1$ $14^x+2^x13^y=2^{z+x}$ $2^{z+x}\equiv 1$ (mod 13) has solution only for $z+x\equiv 0$ (mod 12) ,so $z+x \equiv 0$ (mod 3) ==> $x=3x_1$ $13^y=2^{3z_1}-7^{3x_1}$ ===> $2^{z_1}-7^{x_1}=13^{y_1}$ $13^y=2^{3z_1}-7^{3x_1}$ has solution only if $2^{z_1}-7^{x_1}=13^{y_1}$ has solution for some $y_1$ Set $z_1=3,x_1=1$ - second answer ===> $z=9,x=3$ : $2^9-7^3=169=13^2$ , so $(3,2,9)$ - third answer Set $z_1=9,x_1=3$ - third answer ===>$z=27,x=9$ but $2^{27}-7^9=93864121=169*199*2791$ - not solution. Other $(3x_1,y,3z_1)$ can not be answer, because $(x_1,y_1,z_1)$, where $y_1$ - some natural, cannot be answer.

yes there are this three solution but your proof has some mistakes

dangerousliri wrote: yes there are this three solution but your proof has some mistakes Can you show me my mistakes,please?

RagvaloD wrote: If $y=0$: $7^x+1=2^z$: two obvious solutions : $(0,0,1),(1,0,3)$ And I don`t know are there other solutions first here how do you now this are only this two. RagvaloD wrote: Set $z_1=3,x_1=1$ - second answer ===> $z=9,x=3$ : $2^9-7^3=169=13^2$ , so $(3,2,9)$ - third answer Set $z_1=9,x_1=3$ - third answer ===>$z=27,x=9$ but $2^{27}-7^9=93864121=169*199*2791$ - not solution. Other $(3x_1,y,3z_1)$ can not be answer, because $(x_1,y_1,z_1)$, where $y_1$ - some natural, cannot be answer. second why you just take $z_1=3,x_1=1$ and $z_1=9,x_1=3$ what about other options why can't satisfied

$7^x+1=2^z$ Let $z\geq 4$ $7^x+1\equiv 0 $(mod 16) ,but $7^x+1\equiv 2,8$(mod 16) for every $x$, so $z<4$ From solution we can find, that $(x,y,z)$ can be answer only if $x=3x_1,z=3z_1$, and $(x_1,y_1,z_1)$ is solution for some $y_1\leq y-1$ And by induction , we can prove, that $(x,y,z) $can be answer only if $x=3^tx_t,z=3^tz_t$ and $(x_t,0,z_t)$ is solution. So all possible answers are $(3^t,y,3^{t+1})$ (if $x_t=1,z_t=3$) for some $t$, or $(0,y,3^t)$ (if $x_t=0,z_t=1$) Case 1 :$2^{3^{t+1}}-7^{3^t}$ is divisible $2^{27}-7^9=169*199*2791$ for $t>1$, so $t=0,1$ Case 2: $2^{3^{t}}-7^0$ is divisible 7 for $t>0$, so $t=0$

RagvaloD wrote: So all possible answers are $(3^t,y,3^{t+1})$ (if $x_t=1,z_t=3$) for some $t$, or $(0,y,3^t)$ (if $x_t=0,z_t=1$) why can't be $x_t=2$ or $x_t=7$ ... etc why?

dangerousliri wrote: RagvaloD wrote: So all possible answers are $(3^t,y,3^{t+1})$ (if $x_t=1,z_t=3$) for some $t$, or $(0,y,3^t)$ (if $x_t=0,z_t=1$) why can't be $x_t=2$ or $x_t=7$ ... etc why? By induction. If $y>0$ and $7^x+13^y=2^z$ than there are such $(x_1,y_1,z_1)$ $x=3x_1,y_1<y,z=z_1$ for which $2^{z_1}=7^{x_1}+13^{y_1}$. We must find such triples until $y>0$. For $y=0$ we find all solutions.

i still don't get it anyway here it is my solution: we see that we can suppose $x,y,z$ are non-negative integers If $x=0$ we have $1+13^y=2^z$ if $z\geq 2$ we have $2^z=1+13^y\equiv 2(mod4)$ which is impossible so $z=0$ or $z=1$ for $z=0$ it's impossible and if $z=1$ we have $y=0$ so one solution is $(0,0,1)$ if $y=0$ we have $7^x+1=2^z$ if $x=0$ we have $z=1$ which alread find it and if $x=1$ we have $z=3$ so $(1,0,3)$ is also solution now if $x\geq 2$ from Zsigmondy exist prime number $p\neq 2$ such that $p|7^x+1=2^z$ which is contradiction now we consider $x,y,z$ are positive integers let be $x,y,z$ solution with minimal sum $x+y+z$. $2^z=7^x+13^y\equiv (-1)^y(mod7)$ and since $2^z\equiv 1,2,4$ we have that $y$ is even and $z$ is divisible by $3$ hence $z=3z_1$ $2^{z+x}=14^x+2^x\cdot 13^y\equiv 1(mod13)$ which is possible only if $z+x$ is divisible by $12$ now since $z$ is divisible by $3$ also $x$ is divisible by $3$ so $x=3x_1$ so we can write $13^y=(2^{z_1}-7^{x_1})(4^{z_1}+2^{z_1}\cdot 7^{x_1}+49^{x_1})$ so exist non-negative integer $y_1\leq y-1$ such that $13^{y_1}=2^{z_1}-7^{x_1}\Rightarrow 7^{x_1}+13^{y_1}=2^{z_1}$ if $y_1\neq 0$ we have $(x_1,y_1,z_1)$ is solution of equation $7^x+13^y=2^z$ and $x_1+y_1+z_1<3x_1+y+3z_1=x+y+z$ which contradicts minimality of $x+y+z$ hence $y_1=0$ so we have $7^{x_1}+1=2^{z_1}$ and this equation already solve it which solution are $(0,1)$ and $(1,3)$ since $x_1,z_1$ are positive integers it remains only $x_1=1$ and $z_1=3$ where we have $x=3$ and $z=9$ from here we find $y=2$ so $(3,2,9)$ is also solution. So all solution of equation are $(x,y,z)=(0,0,1),(1,0,3),(3,2,9)$