<s>[INST] What is the general solution of this equation :$2^x 3^y+1=7^z$ with $x, y , z$ are integers? I have got these triplet solution  $(x,y,z)=(1,1,1),(4,1,2)$ for this equation:

$$2^x 3^y+1=7^z$$ 

with  $x, y , z$ are integers, But i can't get general solution of it, I have attempted to use Gausse theorem for the solution of $ ax+by= c $, with $a, b, c$ are integers but my problem i can't transfer the titled equation to that of Gausse as a linear form, any way ?
 [/INST] Clearly $z>0$, since $2^x3^y+1>1$, so $7^z-1\in \mathbb{Z}$ so $x,y\in  \mathbb{N}$.
If $y\geq 2$ then $1\equiv _9 7^z$. Since ord$_9(7) = 3$ we have $3\mid z$ so $z=3t$.
Now we can write: $$2^x3^y = (7^3-1)\Big((7^3)^{t-1}+\ldots+7^3+1\Big)$$
Since $7^3-1 = 19\cdot 9\cdot 2$ we see this is impossible, so $y\leq 1$ or $y=1$. 
Can you finish?

Added:
Now we have: $$2^{x-1} = 7^{z-1}+\ldots+7^2+7+1$$
Say $x>1$, then $z$ is even so $z=2s$ and $$2^x3 = (7^s-1)(7^s+1)$$ 
Since factors on right differ for $2$ at most one is divisible by $4$.  


*

*$7^s-1 = 1$ and $7^s+1 = 2^{x}3$...

*$7^s-1 = 2$ and $7^s+1 = 2^{x-1}3$...

*$7^s-1 = 3$ and $7^s+1 = 2^{x}$...

*$7^s-1 = 6$ and $7^s+1 = 2^{x-1}$...

*$7^s+1 = 1$ and $7^s-1 = 2^{x}3$...

*$7^s+1 = 2$ and $7^s-1 = 2^{x-1}3$...

*$7^s+1 = 3$ and $7^s-1 = 2^{x}$...

*$7^s+1 = 6$ and $7^s-1 = 2^{x-1}$...

</s>