Notice that the RHS is one more than a positive square, so it cannot be a square meaning $x \ne y$. WLOG suppose $x > y$. Expanding, \[2^{x + y} - 2^x - 2^y = 2^{2^z}\] \[2^y(2^{x} - 2^{x - y} - 1) = 2^{2^z}\] Since $2^{x} - 2^{x - y} - 1$ is odd, it must equal one otherwise it would have a nontrivial odd divisor, which the RHS can obviously not have. If $2^{x} - 2^{x - y} - 1 = 1$, \[2^x - 2^{x - y} = 2\] \[2^{x - y}(2^{y} - 1) = 2\] Since $x - y \ge 1$, the only solution to this equation is $(x, y) = (2, 1)$. However, this would require $z = 0$ which is not a natural number, so the equation has no solutions.

It is easy to see that for a natural number $a$, $2^a-1 \equiv 1 \pmod 3$ if $a$ is odd and $2^a-1 \equiv 0 \pmod 3$ if $a$ is even. Thus $(2^x-1)(2^y-1) \equiv 0 \pmod 3$ is any of $x,y$ are even or $\equiv 1 \pmod 3$ if both are odd. Thus $\text{LHS } \not \equiv 2 \pmod 3$. Now the RHS is $2^{2^z}+1 = 2 + (2^{2^z}-1) = 2 +\left((2^2)^{2^{z-1}}-1\right) = 2+ \left(4^{2^{z-1}}-1\right) \equiv 2 \pmod 3$ since $3|\left(4^a-1\right)$ for any natural number $a$. Thus $\text{LHS } \not \equiv 2 \pmod 3$ but $\text{RHS } \equiv 2 \pmod 3$. Hence no solutions.

it turns to $2^{x+y}-2^x-2^y=2^{2^z}$ if $x>y$ , then $2^y(2^x-2^{x-y}-1)=2^{2^z}$ so , $2^x-2^{x-y}-1=1 \implies 2^{x-y}(2^y-1)=2$ so , $2^y=2$ . so , y=1 ,x=2. clearly , this doesn't yield any solution similarly , no solution for $x<y$ just note that ,for $x=y$ ,$x=1$ doesn't yield any solution. if $x>1$ , $2^{x+1}(2^{x-1}-1)=2^{2^z}$. so , $2^{x-1}=2$. clearly , this gives no solution. so , NO SOLUTIONS at all

Potla wrote: Find all natural numbers $x,y,z$ such that \[(2^x-1)(2^y-1)=2^{2^z}+1.\] For the sake of overkill, write \[(2^x-1)(2^y-1)=2^{2^z}+1 \\ \iff (2^{2^z}-1)(2^x-1)(2^y-1)=2^{2^{z+1}}-1\] and apply Zsigmondy's Theorem. (Oh, don't forget to verify the special cases )

hyperbolictangent wrote: Notice that the RHS is one more than a positive square, so it cannot be a square meaning $x \ne y$. WLOG suppose $x > y$. Expanding, \[2^{x + y} - 2^x - 2^y = 2^{2^z}\]\[2^y(2^{x} - 2^{x - y} - 1) = 2^{2^z}\] Since $2^{x} - 2^{x - y} - 1$ is odd, it must equal one otherwise it would have a nontrivial odd divisor, which the RHS can obviously not have. If $2^{x} - 2^{x - y} - 1 = 1$, \[2^x - 2^{x - y} = 2\]\[2^{x - y}(2^{y} - 1) = 2\]Since $x - y \ge 1$, the only solution to this equation is $(x, y) = (2, 1)$. However, this would require $z = 0$ which is not a natural number, so the equation has no solutions. $2^{x}-2^{x-y}-1$ is not odd when $x=y$

Haven't seen this solution here. Note that this assumes "natural numbers" means "$\geq 1$". Suppose $x\geq 2$. Then $2^x-1\equiv 3\pmod 4$, which implies that there exists some prime $p\equiv 3\pmod 4$ such that $p\mid 2^x-1$. But then $p$ divides $2^{2^z} + 1 = (2^{2^{z-1}})^2 + 1$, which is a contradiction. Therefore $x=1$, and similarly $y=1$. But this pair $(x,y)$ does not work, since then \[ 2^{2^z} = (2^1-1)(2^1-1) - 1 = 1\cdot 1 - 1 = 0. \]So there are no solutions.

Haven't seen this too, we know $2^a-1$ never divide $2^b+1$ where a,b are positive integers>2, case 1: x=1 so y=1 so no solns here case 2 x=2 not possible as 3 does not divide RHS, SO no solns