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 ?

As Aqua notes, we must have $z\gt0$, so let's write $z=c+1$. Then
$$2^x3^y=7^{c+1}-1=(7-1)(1+7+7^2+\cdots+7^c)$$
It follows that $x,y\gt0$ as well, so let's write $x=a+1$ and $y=b+1$. We now have
$$2^a3^b=1+7+7^2+\cdots+7^c$$
If $b\gt0$, then, since $7\equiv1$ mod $3$, the number of terms on the right hand side must be a multiple of $3$, in which case we have
$$1+7+7^2+\cdots+7^c=(1+7+7^2)(1+7^3+7^6+\cdots+7^{3k})$$
for some $k\ge0$. But $1+7+7^2=57=3\cdot19$, and $19$ is not divisible by $2$ or $3$.  So we must have $b=0$.
If $a=0$, we have the solution $(x,y,z)=(1,1,1)$. If $a\gt0$, we must have an even number of terms in $1+7+7^2+\cdots+7^c$, so that
$$1+7+7^2+\cdots+7^c=(1+7)(1+7^2+7^4+\cdots+7^{2k})$$
for some $k\ge0$. Thus $a\gt0$ implies $a=3+a'$ for some $a'\ge0$, with $2^{a'}=1+7^2+7^4+\cdot+7^{2k}$. Now if $a'\gt0$, we must have an even number of terms in $1+7^2+7^4+\cdots+7^{2k}$, in which case we have
$$1+7^2+7^4+\cdots+7^{2k}=(1+7^2)(1+7^4+7^8+\cdots+7^{4k'})$$
for some $k'\ge0$. But $1+7^2=50=2\cdot5^2$ is not a power of $2$. So we must have $a'=0$, and this leaves $(x,y,z)=(4,1,2)$ as the only other solution.
Remark: The prime $19$ plays the same role here as it does in Aqua's answer, but it is obtained in a somewhat different manner (with a mod $3$ argument rather than a mod $9$ observation). The approach here can be modified to solve the equation $2^x3^y5^z+1=31^w$.
