I shall prove the same using modular arithmetic keeping arithmetic simple
We have $2^5 = 32 \equiv -15 \pmod {47}$
$=> 2^{10} \equiv (-15) ^2 \pmod {47}$
$=> 2^{10} \equiv 225 \pmod {47}$
$=> 2^{10} \equiv -10 \pmod {47}$ (as 225 = 5 * 47 -10)
$=> 2^{20} \equiv (-10)^2 \pmod {47}$
$=> 2^{20} \equiv 100 \pmod {47}$
$=> 2^{20} \equiv 6 \pmod {47}$ as 100 =2 * 47 + 6
$=> 2^{23} \equiv 6 * 2^3 \pmod {47}$
$=> 2^{23} \equiv 48 \pmod {47}$
$=> 2^{23} \equiv 1 \pmod {47}$
$=> 2^{23} -1 \equiv 0 \pmod {47}$
Hence $2^{23} -1$ is divisible by 47
