To solve the sum \( S = 1 + 2 + 2^2 + 2^3 + \ldots + 2^{2023} \), we recognize that this is a geometric series with the first term \( a = 1 \) and the common ratio \( r = 2 \).

The sum of the first \( n \) terms of a geometric series is given by:
\[ S_n = a \frac{r^n - 1}{r - 1} \]

For this series, \( a = 1 \), \( r = 2 \), and \( n = 2024 \) (since the series goes up to \( 2^{2023} \), there are 2024 terms in total).

So, the sum \( S \) is:
\[ S = 1 \cdot \frac{2^{2024} - 1}{2 - 1} = 2^{2024} - 1 \]

Therefore, the sum \( S \) is:
\[ S = 2^{2024} - 1 \]

For the second part, it seems to be asking for the comparison between \( S^2 \), \( S \sqrt{S} \), and \( S_{2024} \). However, without additional context or clarification, it's challenging to interpret the exact meaning. If we assume \( S_{2024} \) refers to the sum of the series up to \( 2^{2024} \), then:

\[ S_{2024} = 1 + 2 + 2^2 + 2^3 + \ldots + 2^{2024} \]

Using the same formula for the sum of a geometric series:
\[ S_{2024} = 1 \cdot \frac{2^{2025} - 1}{2 - 1} = 2^{2025} - 1 \]

So, we have:
\[ S_{2024} = 2^{2025} - 1 \]

Comparing \( S \) and \( S_{2024} \):
\[ S = 2^{2024} - 1 \]
\[ S_{2024} = 2^{2025} - 1 \]

Clearly, \( S_{2024} \) is much larger than \( S \).

For \( S^2 \) and \( S \sqrt{S} \):
\[ S^2 = (2^{2024} - 1)^2 \]
\[ S \sqrt{S} = (2^{2024} - 1) \sqrt{2^{2024} - 1} \]

These expressions are more complex and would require further simplification or numerical methods to compare directly. However, it is evident that \( S_{2024} \) is exponentially larger than both \( S^2 \) and \( S \sqrt{S} \).