When $p=2$, then any even number $n$ will satisfy the requirement.

When $p > 2$, by Fermat little theorem, we have $2^{p-1}\equiv 1\pmod{p}$. This means that we have $2^{k(p-1)}\equiv 1\pmod{p}$ for any integer $k$.

Now, if we can show that there exist infinitely many $k$ such that $k(p-1)\equiv 1\pmod{p}$, then letting $n=k(p-1)$ will yield $$2^n\equiv 1\equiv n\pmod{p} \implies p\mid (2^n-n)$$

Indeed, there exist infinitely many such $k$ because as long as $k\equiv -1\pmod{p}$, we will have $k(p-1)\equiv 1\pmod{p}$.