If you accept that 0 is not a natural number, then there is a very simple answer to your question: take $P$ to be all numbers whose expansions base 4 contain only digits 0 and 1 and $Q$ to contain only 0 and 2. Then $P\cap Q=\{0\}$, which we have boldly excluded.

Also, both sets have the lowest possible asymptotic density of order $1/\sqrt n$, which is kinda nice.

If you accept that 0 is not a natural number, then there is a very simple answer to your question: take $P$ to be all numbers whose expansions base 4 contain only digits 0 and 1 and $Q$ to contain only digits 0 and 2. Then $P\cap Q=\{0\}$, which we have boldly excluded.

Also, both sets have the lowest possible asymptotic density of order $1/\sqrt n$, which is kinda nice.