$z \ge 3$, then $2^z - 1 \vdash 3$, but it is possible only if $z$ is even. Let $z = 2c$ for some nonnegative integer $c$. Then $4^c - 3^a = 1$. If $a = 1$, we find $c = 1$ and obtain $x = 1, z = 2, y = 1$. If $a > 1$, then $4^c = 3^a + 1$ has remainder 1 when divided by 9. The power of 4 has this remainder only if $c$ is divisible by 3. But in this case $4^3 \equiv 1 \pmod 7$ and hence $4^c \equiv 1 \pmod 7$. It follows that $3^a = 4^c - 1 \vdash 7$, which is impossible.