For which rational $ x$ is the number $ 1 \plus{} 105 \cdot 2^x$ the square of a rational number?
To determine for which rational \( x \) the number \( 1 + 105 \cdot 2^x \) is the square of a rational number, we start by setting up the equation:

\[ 1 + 105 \cdot 2^x = n^2 \]

where \( n \) is a rational number. Since \( 1 + 105 \cdot 2^x \) must be an integer, \( 2^x \) must also be an integer, implying that \( x \) must be an integer.

1. **Check for \( x = 0 \):**
   \[ 1 + 105 \cdot 2^0 = 1 + 105 = 106 \]
   Since 106 is not a perfect square, \( x = 0 \) is not a solution.

2. **General Case:**
   Let \( 1 + 105 \cdot 2^x = n^2 \). Then:
   \[ n^2 - 1 = 105 \cdot 2^x \]
   \[ (n-1)(n+1) = 105 \cdot 2^x \]

   Since \( n-1 \) and \( n+1 \) are two consecutive even numbers, one of them must be divisible by 4. We need to find pairs \((a, b)\) such that:
   \[ a \cdot b = 105 \cdot 2^x \]
   where \( a = n-1 \) and \( b = n+1 \).

3. **Factorization of 105:**
   The prime factorization of 105 is:
   \[ 105 = 3 \cdot 5 \cdot 7 \]

   We need to find pairs \((a, b)\) such that \( a \cdot b = 105 \cdot 2^x \) and \( a \) and \( b \) differ by 2. We will check the possible values of \( x \).

4. **Positive Solutions:**
   - For \( x = 3 \):
     \[ 1 + 105 \cdot 2^3 = 1 + 105 \cdot 8 = 1 + 840 = 841 \]
     \[ 841 = 29^2 \]
     Thus, \( x = 3 \) is a solution.

   - For \( x = 4 \):
     \[ 1 + 105 \cdot 2^4 = 1 + 105 \cdot 16 = 1 + 1680 = 1681 \]
     \[ 1681 = 41^2 \]
     Thus, \( x = 4 \) is a solution.

5. **Negative Solutions:**
   - For \( x = -4 \):
     \[ 1 + 105 \cdot 2^{-4} = 1 + 105 \cdot \frac{1}{16} = 1 + \frac{105}{16} = \frac{16 + 105}{16} = \frac{121}{16} = \left(\frac{11}{4}\right)^2 \]
     Thus, \( x = -4 \) is a solution.

   - For \( x = -6 \):
     \[ 1 + 105 \cdot 2^{-6} = 1 + 105 \cdot \frac{1}{64} = 1 + \frac{105}{64} = \frac{64 + 105}{64} = \frac{169}{64} = \left(\frac{13}{8}\right)^2 \]
     Thus, \( x = -6 \) is a solution.

   - For \( x = -8 \):
     \[ 1 + 105 \cdot 2^{-8} = 1 + 105 \cdot \frac{1}{256} = 1 + \frac{105}{256} = \frac{256 + 105}{256} = \frac{361}{256} = \left(\frac{19}{16}\right)^2 \]
     Thus, \( x = -8 \) is a solution.

Conclusion:
The values of \( x \) for which \( 1 + 105 \cdot 2^x \) is the square of a rational number are \( x = 3, 4, -4, -6, -8 \).

The final answer is \( \boxed{ x = 3, 4, -4, -6, -8 } \).