If none of $x, y, z$ is nil, then the left term of the equation is even, so it cannot be equal to $1023$. Hence, at least one of $x, y, z$ is nil.
From $2^x = 2^y + 2^z + 1023$ follows that $x \ge 11$, therefore $y = 0$ or $z = 0$.
The case $z = 0$ yields $2^x - 2^y = 1024$, so $2^y \cdot (2^{x-y} - 1) = 2^{10}$, (1). From $2^x - 2^y > 0$ follows $x > y$, so $2^{x-y} - 1$ is odd. From (1), $2^y = 2^{10}$ and $2^{x-y} - 1 = 1$, whence $y = 10$ and $x = 11$.
A similar argument works for $y = 0$, so the solutions are $x = 11, y = 10, z = 0$ and $x = 11, y = 0, z = 10$.