Find all natural numbers $ n$ for which $ 2^n\minus{}n^2$ is divisible by $ 7$.
We are tasked with finding all natural numbers \( n \) for which \( 2^n - n^2 \) is divisible by \( 7 \). This can be expressed as:
\[ 2^n \equiv n^2 \pmod{7} \]

1. **Identify the periodicity of \( 2^n \mod 7 \):**
   Since \( 2^3 \equiv 1 \pmod{7} \), the sequence \( 2^n \mod 7 \) is periodic with a period of 3. We can list the residues:
   \[
   \begin{aligned}
   2^0 &\equiv 1 \pmod{7}, \\
   2^1 &\equiv 2 \pmod{7}, \\
   2^2 &\equiv 4 \pmod{7}, \\
   2^3 &\equiv 1 \pmod{7}, \\
   2^4 &\equiv 2 \pmod{7}, \\
   2^5 &\equiv 4 \pmod{7}, \\
   &\vdots
   \end{aligned}
   \]
   Thus, the residues repeat every 3 terms: \( 1, 2, 4 \).

2. **Identify the periodicity of \( n^2 \mod 7 \):**
   Since \( 7 \) is prime, the sequence \( n^2 \mod 7 \) is periodic with a period of 7. We can list the residues:
   \[
   \begin{aligned}
   0^2 &\equiv 0 \pmod{7}, \\
   1^2 &\equiv 1 \pmod{7}, \\
   2^2 &\equiv 4 \pmod{7}, \\
   3^2 &\equiv 2 \pmod{7}, \\
   4^2 &\equiv 2 \pmod{7}, \\
   5^2 &\equiv 4 \pmod{7}, \\
   6^2 &\equiv 1 \pmod{7}, \\
   &\vdots
   \end{aligned}
   \]
   Thus, the residues repeat every 7 terms: \( 0, 1, 4, 2, 2, 4, 1 \).

3. **Combine the periodicities:**
   The combined periodicity of \( 2^n \mod 7 \) and \( n^2 \mod 7 \) is the least common multiple of 3 and 7, which is 21. Therefore, we need to check \( n \) from 0 to 20.

4. **List the residues and find matches:**
   \[
   \begin{aligned}
   2^n \mod 7 &: 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4 \\
   n^2 \mod 7 &: 0, 1, 4, 2, 2, 4, 1, 0, 1, 4, 2, 2, 4, 1, 0, 1, 4, 2, 2, 4, 1
   \end{aligned}
   \]
   We find the values of \( n \) for which \( 2^n \equiv n^2 \pmod{7} \):
   \[
   \begin{aligned}
   n = 2 &: 2^2 \equiv 4 \pmod{7}, \quad 2^2 \equiv 4 \pmod{7} \\
   n = 4 &: 2^4 \equiv 2 \pmod{7}, \quad 4^2 \equiv 2 \pmod{7} \\
   n = 5 &: 2^5 \equiv 4 \pmod{7}, \quad 5^2 \equiv 4 \pmod{7} \\
   n = 6 &: 2^6 \equiv 1 \pmod{7}, \quad 6^2 \equiv 1 \pmod{7} \\
   n = 10 &: 2^{10} \equiv 2 \pmod{7}, \quad 10^2 \equiv 2 \pmod{7} \\
   n = 15 &: 2^{15} \equiv 4 \pmod{7}, \quad 15^2 \equiv 4 \pmod{7}
   \end{aligned}
   \]

Thus, the natural numbers \( n \) such that \( 2^n \equiv n^2 \pmod{7} \) are:
\[ n \equiv 2, 4, 5, 6, 10, 15 \pmod{21} \]

The final answer is \( \boxed{ n \equiv 2, 4, 5, 6, 10, 15 \pmod{21} } \).