Solving $2^{x-z} + 2^{y-z} = 1$ Suppose we have an equation $2^{x-z} + 2^{y-z}  = 1$,where $x, y, z$ are integers from $0$  to $9$.  $x, y, z$ can have same values. I guess $x$ and $y$ should have $9$ possible values like $x=y=0$ and $z=1$ and so on . Or should it have more values? 

Without loss of generality let $x \leq y$ ad write $1+2^{y-x}=2^{z-x}$. We can show that $x=y$ is necessary. Assume that opposite false, i.e. $y>x$, then left side is odd integer, therefore right side must be odd integer as well, in other words $z-x=0$. But that would mean $1+2^{y-x}=1$, or $2^{y-x}=0$, impossible. Therefore $x=y$ and so $2=2^{z-x}$, i.e. $z-x=1$. Thus we have found all solutions $x=y=z-1$ (subjected to the set constrains).
Another way to look this (perhaps more intuitive way) is to look at binary representations of $2^x+2^y=2^z$. Since all three numbers $2^x,2^y,2^z$ have just one $1$ in its binary representation, the sum on the left must make one $1$. The only way for this is if $x=y$ again.
