Solve for positive integers $x,y,z$ Find positive integers $x,y,z$ such that:
$$(x+y)(1+xy)=2^z.$$
I have come across a few solutions for this equation: $(1,1,2)$  and $(1,3,4),(3,5,7),...,(2^k-1,2^k+1,2^{3k}+1)$.
I think it has an infinite number of solutions but I don't know how to solve for all of them. 
 A: One way of attacking the problem is to realize that if the product is a power of 2, then each factor must be one, so we have for some integers $r,s \ge 0$
$$x+y=2^r, 1+xy=2^s$$
and this leads to a quadratic equation:
$$y=2^r-x \rightarrow 1+x(2^r-x)=2^s \rightarrow x^2-2^rx +2^s-1=0$$
which has the solutions:
$$x_{1,2}=2^{r-1} \pm \sqrt{2^{2r-2}-2^s+1}.$$
So it remains to find out for what values of $r,s$ the term $2^{2r-2}-2^s+1$ is a complete square.
You can easily see that $r=0$ does not yield positive solitions for $x$ and $y$.
The series of solutions you found corresponds to $s=2r-2$, where the term under the square root is just $1$ and we get
$$x=2^{r-1}-1, y=2^{r-1}+1, z= r+s=3r-2$$
which corresponds to your solution with $k:=r-1$.
Another series of solutions exists for $s=r$, as then the critical term becomes 
$$2^{2r-2}-2^r+1=2^{2r-2}-2\times2^{r-1}+1=(2^{r-1}-1)^2,$$
and we get with
$$x=1, y=2^r-1, z=r+s=2r$$
another series of solutions. The $(1,1,2)$ you found is the starting value of the series for $r=1$, and it continues with $(1,3,4)$ (this value comes up on both series'), then $(1,7,6),(1,15,8)$, a.s.o.
Are there more solutions? I don't know.
It is possible to further look at $2^{2r-2}-2^s+1$, but non of the techniques I tried lead to a result either way.
