Solving $2^x+2^y+1=3^z$ in integers I have reasons to believe that there should be an elementary, relatively simple way to find all solutions of the equation in the title in positive integers $x>y$ and $z$. Any ideas?

I think I was eventually able to recover (?) the elementary solution mentioned in the literature. I will show below that the only solutions in positive integers $x,y,z$ are $x=4,y=6,z=4$, $x=6,y=4,z=4$, and $x=y=z=2$.
Reducing  $2^x+2^y+1=3^z$ modulo $3$, we conclude that both $x$ and $y$ are even; say, $x=2a$ and $y=2b$, leading to $4^a+4^b+1=3^z$. Reducing now modulo $4$ we see that $z$ is even, too, and the substitution $z=2c$ gives $4^a+4^b+1=9^c$.
Assume without loss of generality that $a\le b$. If $a=1$, then $4^b+5=9^c$  whence $b=1$ and therefore also $c=1$, as otherwise reducing modulo $16$ we would get $9^c\equiv 5\pmod{16}$, which is impossible (the order of $9$ modulo $16$ is $2$). Thus $a\ge 2$, and reducing $4^a+4^b+1=9^c$ modulo $16$ we obtain $9^c\equiv 1\pmod{16}$, whence $c$ is even; we write $c=2u$ to get $4^a+4^b+1=81^u$.
We know that $a\ge 2$. If, indeed, $a=2$, then $4^b+17=81^u$, whence $(9^u-2^b)(9^u+2^b)=17$, leading to $u=1,b=3$; that is, $x=z=4$, $y=6$.
We are thus left with the case where $b\ge a\ge 3$. Reducing $4^a+4^b+1=81^u$ modulo $64$ and observing that the order of $81$ modulo $64$ is $4$, we conclude that $u$ is divisible by $4$, whence $4^a+4^b+1=3^{16v}$, $u=4v$. Reducing then modulo $85$ (which is a divisor of $3^{16}-1$), we get $4^a+4^b\equiv 0\pmod{85}$. Dividing through by $4^a$ gives $4^{b-a}\equiv -1\pmod{85}$. However, there is no power of $4$ congruent to $-1$ modulo $85$, as one can easily verify (the order of $4$ modulo $85$ is $4$). This shows that there are no solutions with $a,b\ge 3$.
