74th Romanian Mathematical Olympiad

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 $x=11,y=0,z=10$.