To solve the equation \( 2^{x + y} = 2^{2x - y} = \sqrt{8} \), we will first express \(\sqrt{8}\) in terms of base 2.

Step 1: Simplifying \(\sqrt{8}\)
- We know that \(\sqrt{8} = 2^{3/2}\).
- Therefore, we can rewrite the equation as:
\( 2^{x + y} = 2^{3/2} \) and \( 2^{2x - y} = 2^{3/2} \)

Step 2: Setting Up the Equations
- From \( 2^{x + y} = 2^{3/2} \), we have:
\( x + y = \frac{3}{2} \) (Equation 1)
- From \( 2^{2x - y} = 2^{3/2} \), we have:
\( 2x - y = \frac{3}{2} \) (Equation 2)

Step 3: Solving the System of Equations
- We now solve the system of equations:
1. **From Equation 1:**
- Rearranging gives us:
\( y = \frac{3}{2} - x \)
2. **Substituting into Equation 2:**
- Replace \(y\) in Equation 2:
\( 2x - \left(\frac{3}{2} - x\right) = \frac{3}{2} \)
- Simplifying this gives:
\( 2x - \frac{3}{2} + x = \frac{3}{2} \)
\( 3x - \frac{3}{2} = \frac{3}{2} \)
\( 3x = 3 \)
\( x = 1 \)

Step 4: Finding \(y\)
- Using the value of \(x\) in Equation 1:
\( y = \frac{3}{2} - 1 = \frac{1}{2} \)

Conclusion
- Thus, the values of \(x\) and \(y\) are:
- \( x = 1 \)
- \( y = \frac{1}{2} \)