The only solutions to the Diophantine equation $O_x=2^y-1$ with $x$ prime are $(2,1)$, $(3,2)$ and $(5,4)$ (assuming O_2=1 is the empty product). Proof:

Suppose $x\ge7$. The number $O_x$ is divisible by $21$ but not by $63$ since $O_x$ is squarefree. Hence $2^y-1$ is divisible by $21$, but not by $63$, so $O(x)\equiv21,42 (\mod 63)$.

$2^x\mod 63$ is $1, 2, 4, 8, 16, 32$ for $x\mod6=0, 1, 2, 3, 4, 5$, respectively. Hence if $x$ is an integer, then $2^x-1\equiv0, 1, 3, 7, 15, 31$ mod 63.

This gives a contradiction, so $x\lt7$. The remaining cases can be done by hand.

The only solutions to the Diophantine equation $O_x=2^y-1$ with $x$ prime are $(2,1)$, $(3,2)$ and $(5,4)$ (assuming $O_2=1$ is the empty product). Proof:

Suppose $x\ge7$. The number $O_x$ is divisible by $21$ but not by $63$ since $O_x$ is squarefree. Hence $2^y-1$ is divisible by $21$, but not by $63$, so $O(x)\equiv21,42 (\mod 63)$.

$2^x\mod 63$ is $1, 2, 4, 8, 16, 32$ for $x\mod6=0, 1, 2, 3, 4, 5$, respectively. Hence if $x$ is an integer, then $2^x-1\equiv0, 1, 3, 7, 15, 31$ mod 63.

This gives a contradiction, so $x\lt7$. The remaining cases can be done by hand.