Powers of $2$ and the products of initial odd primes

Question

NOTATION: $O_x$ -- the product of all odd primes $\le x$.

E.g. $O_7=3\cdot 5\cdot 7 = 105$.

QUESTION: Are the three ordered pairs $\ (d\ p)=(1\ 3)\ \ (2\ 3)\ \ (4\ 5)\ $ the only solutions of the equation: $$|O_p-2^d|=1$$ in natural numbers $d$, and odd primes $p$?

(I don't know an answer).

MOTIVATION: Let $\ s\ $ be a prime just after $\ p$, so that $\ p<s$. If $$|O_p-2^d|\ne 1$$ then $$|O_p-2^d|\ge s$$ Furthermore, sometimes $\ s\ $ can be quite a bit larger than $\ p$.

ACKNOWLEDGEMENT: Bjørn Kjos-Hanssen has provided one of the above solutions (see his comment below).

Answer 1

There is no solution for $O_n=2^d-1$ with $n \geq 7$.

If 5 divides $2^d-1$ and 7 divides $2^d-1$, then 9 divides $2^d-1$. [Because 4 divides $d$ and 3 divides $d$; 6 divides $d$ and hence $2^{d}-1$ is divisible by 9.]

Answer 2

Let's complete the answer following The Masked Avenger: modulo $7$ the powers of $2$ are $2,4,1$ and then repeat cyclically, so that $1+2^d$ is never divisible by $7$.

Answer 3

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.