Diophantine triplet

$2^x + 1 = 7^y + 2^z \\ 2^x - 2^z = 7^y - 1 \\ 2^z \cdot (2^{x-z} - 1) = (7 - 1) \cdot \sum\limits_{i=0}^{y-1} 7^i \\ 2^z \cdot (2^{x-z} - 1) = 2 \cdot 3 \cdot \sum\limits_{i=0}^{y-1} 7^i \\ 2^z \cdot (2^{x-z} - 1) = 2 \cdot (2^2 - 1) \cdot \sum\limits_{i=0}^{y-1} 7^i \\$ As there is factor 3 at right side, so $x-z$ must be even. Let $x-z = 2k$ , then $\\ 2^z \cdot (2^{x-z} - 1) = 2^z \cdot (4^k - 1) \\ = 2^z \cdot (2^2 - 1) \cdot \sum\limits_{i=0}^{k-1} 4^i \\ \\ \therefore 2^{z-1} \cdot \sum\limits_{i=0}^{k-1} 4^i = \sum\limits_{i=0}^{y-1} 7^i \\$

• if $y = 1$ , then $z= 1, x = 3$

• if $y = 2$ , then $z= 4, x = 6$

• any else?