Elementary Diophantine equation Solve $(x+y)(xy+1)=2^z$ in positive integres. My attempts is to use $x+y=2^a$, $xy=2^b-1$ and therefore $x,y$ are the roots of the quadratic equation $w^2-2^aw+2^b-1=0$. I try to analyze its dicriminant but it seems to be a dead end...

ETA: Jyrki's answer and comments already says this it turns out, and so he really should get the bounty.
Let $x$ and $y$ be positive integers satisfying $x+y\ge 16$. Then $x+y$ must be a power of $2$. Write then, $x+y=2^a$ for some $a \ge 4$. Then $xy+1$ must also be a power of $2$, and as the strict inequality $xy+1 \ge x+y$ for all such $x,y$, it follows that
$$xy+1=(2^a-y)y+1 =2^b$$ $$= 2^ay -y^2+1,$$ where $b$ is an integer satisfying $b>a$. And so $2^a| (y^2-1)$. So from this it follows that $2^{a-1}$ divides one of $y+1, y-1$ [because $4$ can only divide one of $y+1,y-1$]. Likewise, $2^{a-1}$ divides one of $x-1,x+1$. So from this one can deduce WLOG that $2^{a-1}$ divides $y+1$ and $2^{a-1}$ divides $x-1$, as this is the only way $x$ and $y$ can sum to something that is $0$ mod $2^a$. As $x+y=2^a$ and both $x$ and $y$ are nonnegative integers, it follows that $x$ and $y$ must respectively satisfy either the equations $x=2^{a-1}+1$ and $y=2^{a-1}-1$, or $y=2^a-1$ and $x=1$.
However, for each positive integer $a  \ge 4$, note that if on the one hand either $x$ and $y$ satisfy the respective equations $x=2^{a-1}+1$ and $y=2^{a-1}-1$, then $(x,y,z)$, with $z=3a-2$, is a solution: $$(x+y)(xy+1)=2^a((2^{2a-2}-1)+1) = 2^{3a-2}.$$ If on the other hand $x=1$ and $y=2^a-1$ then $(x,y,z)$, with $z=2a$ is also a solution: $$(x+y)(xy+1)=(1+y)(1+y)$$ $$=2^a×2^a =2^{2a}.$$
