1. We need the remainder when $2021^{2023}$ is divided by $7$.

2. First reduce the base modulo $7$: $2021 \div 7$ Since $7 \times 288 = 2016,$ we get $2021 \equiv 2021-2016 \equiv 5 \pmod{7}.$

   So, $2021^{2023} \equiv 5^{2023} \pmod{7}.$

3. Now simplify powers of $5$ modulo $7$: $5 \equiv -2 \pmod{7}.$ Hence, $5^{2023} \equiv (-2)^{2023} \pmod{7}.$

   But using cyclicity is easier: $5^1 \equiv 5 \pmod{7}$ $5^2 = 25 \equiv 4 \pmod{7}$ $5^3 = 125 \equiv 6 \pmod{7}$ $5^4 = 625 \equiv 2 \pmod{7}$ $5^5 = 3125 \equiv 3 \pmod{7}$ $5^6 \equiv 1 \pmod{7}$

   So the powers repeat with period $6$.

4. Reduce the exponent modulo $6$: $2023 \div 6$ Since $6 \times 337 = 2022,$ we get $2023 \equiv 1 \pmod{6}.$

   Therefore, $5^{2023} \equiv 5^1 \equiv 5 \pmod{7}.$

5. Hence the remainder is $\boxed{5}.$

6. Checking options:

   • A: $1$ ❌
   • B: $2$ ❌
   • C: $5$ ✅
   • D: $6$ ❌

So the correct option is C.