Because $2$ and $47$ are relatively prime numbers, $47$ divides $3^{x} - 2^{y}$ if and only if $47$ divides $2^{4x}(3^{x} - 2^{y})$. But
$$2^{4x}(3^{x} - 2^{y}) = 48^{x} - 2^{4x + y} \equiv 1 - 2^{4x + y} \pmod{47}.$$
Therefore $47$ divides $3^{x} - 2^{y}$ if and only if $2^{4x + y} \equiv 1 \pmod{47}$.

On the other hand, we have $2^{23} \equiv 49^{23} \equiv 7^{46} \equiv 1 \pmod{47}$. We deduce that the prime number $23$ is the order of $2$ modulo $47$. This implies that $2^{4x + y} \equiv 1$ mod $47$ if and only if $23$ divides $4x + y$.