What is the general solution of this equation :$2^x 3^y+1=7^z$ with $x, y , z$ are integers?

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 $4$.

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