Find all natural numbers $n$ for which the following $5$ conditions hold:
$(1)$ $n$ is not divisible by any perfect square bigger than $1$.
$(2)$ $n$ has exactly one prime divisor of the form $4k+3$, $k\in \mathbb{N}_0$.
$(3)$ Denote by $S(n)$ the sum of digits of $n$ and $d(n)$ as the number of positive divisors of $n$. Then we have that $S(n)+2=d(n)$.
$(4)$ $n+3$ is a perfect square.
$(5)$ $n$ does not have a prime divisor which has $4$ or more digits.
1. **Sum of digits and number of divisors:**

   Given \( S(n) + 2 = d(n) \), we need to find \( S(n) \) and \( d(n) \).

   For \( n = 2 \cdot q \cdot p_1 \cdot p_2 \cdots p_s \):
   \[ d(n) = 2^{s+2} \]
   Therefore,
   \[ S(n) = 2^{s+2} - 2 \]

2. **Bounding \( n \):**

   Since \( n < 2 \cdot 10^{3(s+1)} \), the number of digits of \( n \) is at most \( 3s + 4 \). If equality holds, the leading digit is 1, so:
   \[ S(n) \leq 27(s+1) + 1 \]

   Hence,
   \[ 27s + 30 \geq 2^{s+2} \]

   By proving the reverse inequality by induction, we deduce that \( s \leq 5 \).

3. **Modulo 9 condition:**

   Since \( n \equiv 6 \pmod{9} \), we have:
   \[ 2^{s+2} - 2 \equiv 6 \pmod{9} \]

   This is violated by \( s = 2, 3, 4, 5 \). Therefore, we must have \( s = 1 \).

4. **Case \( s = 1 \):**

   Then \( S(n) = 6 \). Since \( n \equiv S(n) \equiv 0 \pmod{3} \), we get \( q = 3 \) and \( n = 6p_1 \).

   With \( n + 3 = m^2 \), we must have \( m = 3(2k+1) \) and so:
   \[ p_1 = 6k^2 + 6k + 1 \]

   This is less than 1000 for \( k \leq 12 \). The corresponding primes are:
   \[ 13, 37, 73, 181, 337, 433, 541, 661, 793, 937 \]

   If \( p_1 \) ends on 1 or 3, then \( 6p_1 \) has last digit 6 or 8, and the sum of digits exceeds 6. The valid solutions are:
   \[ n = 6 \cdot 37 = 222 \]
   \[ n = 6 \cdot 337 = 2022 \]

   Therefore, the only solutions are \( 222 \) and \( 2022 \).

The final answer is \( \boxed{222 \text{ and } 2022} \).