Erdos's argument essentially gives that there is an infinite
arithmetic progression of odd numbers containing
no element of the form power of two minus a prime.

Define

\begin{array}{cccc}
i & a_i & m_i & p_i \\
1 & 1 & 2 & 3 \\
2 & 2 & 4 & 5 \\
3 & 4 & 8 & 17 \\
4 & 8 & 16 & 257 \\
5 & 16 & 32 & 65537 \\
6 & 0 & 64 & 641 \\
7 & 32 & 64 & 6700417 \\
\end{array}

Set $A_0=\{x:x\equiv 1 \pmod{8}\}$,
$A_i=\{x:x\equiv 2^{a_i} \pmod{p_i}\}$ ($1\leq i\leq 7$),
$B=A_0\cap A_1\cap\cdots\cap A_7$.

$B$ is an AP consisting of odd numbers.
Assume that $x\in B$ and $x=2^n-q$ for some odd prime $q$.
For some $1\leq i\leq 7$, we have $n\equiv a_i \pmod{m_i}$,
therefore $x\equiv 2^{a_i}-q \pmod{p_i}$. As $x\in A_i$, we
also have $x\equiv 2^{a_i} \pmod{p_i}$, consequently $q=p_i$.
But this is impossible as $2^n\equiv 0 \pmod{8}$ for $n\geq 3$,
$p_i\equiv 1,3,5 \pmod{8}$ and as $x\in A_0$, $x\equiv 7 \pmod{8}$.