Suppose that $x$ and $y$ are real numbers that satisfy the system of equations

$2^x-2^y=1$
$4^x-4^y=\frac{5}{3}$

Determine $x-y$
1. Let \( 2^x = a \) and \( 2^y = b \). Then the given system of equations becomes:
   \[
   a - b = 1
   \]
   \[
   a^2 - b^2 = \frac{5}{3}
   \]

2. Notice that \( a^2 - b^2 \) can be factored using the difference of squares:
   \[
   a^2 - b^2 = (a - b)(a + b)
   \]

3. Substitute \( a - b = 1 \) into the factored form:
   \[
   (a - b)(a + b) = 1 \cdot (a + b) = \frac{5}{3}
   \]
   Therefore:
   \[
   a + b = \frac{5}{3}
   \]

4. Now we have the system of linear equations:
   \[
   a - b = 1
   \]
   \[
   a + b = \frac{5}{3}
   \]

5. Solve this system by adding the two equations:
   \[
   (a - b) + (a + b) = 1 + \frac{5}{3}
   \]
   \[
   2a = \frac{8}{3}
   \]
   \[
   a = \frac{4}{3}
   \]

6. Substitute \( a = \frac{4}{3} \) back into \( a - b = 1 \):
   \[
   \frac{4}{3} - b = 1
   \]
   \[
   b = \frac{4}{3} - 1 = \frac{4}{3} - \frac{3}{3} = \frac{1}{3}
   \]

7. Now we have \( a = \frac{4}{3} \) and \( b = \frac{1}{3} \). Recall that \( a = 2^x \) and \( b = 2^y \):
   \[
   2^x = \frac{4}{3}
   \]
   \[
   2^y = \frac{1}{3}
   \]

8. Take the logarithm base 2 of both sides to solve for \( x \) and \( y \):
   \[
   x = \log_2 \left( \frac{4}{3} \right)
   \]
   \[
   y = \log_2 \left( \frac{1}{3} \right)
   \]

9. To find \( x - y \):
   \[
   x - y = \log_2 \left( \frac{4}{3} \right) - \log_2 \left( \frac{1}{3} \right)
   \]
   Using the properties of logarithms:
   \[
   x - y = \log_2 \left( \frac{4}{3} \cdot \frac{3}{1} \right)
   \]
   \[
   x - y = \log_2 (4) = 2
   \]

The final answer is \( \boxed{2} \).