Find the number of functions defined on positive real numbers such that $ f\left(1\right) \equal{} 1$ and for every $ x,y\in \Re$, $ f\left(x^{2} y^{2} \right) \equal{} f\left(x^{4} \plus{} y^{4} \right)$. 

$\textbf{(A)}\ 0  \qquad\textbf{(B)}\ 1  \qquad\textbf{(C)}\ 2  \qquad\textbf{(D)}\ 4  \qquad\textbf{(E)}\ \text{Infinitely many}$
1. Given the function \( f \) defined on positive real numbers such that \( f(1) = 1 \) and for every \( x, y \in \mathbb{R} \), \( f(x^2 y^2) = f(x^4 + y^4) \).

2. Let's analyze the functional equation \( f(x^2 y^2) = f(x^4 + y^4) \). 

3. Set \( x = 1 \) and \( y = 1 \):
   \[
   f(1^2 \cdot 1^2) = f(1^4 + 1^4) \implies f(1) = f(2)
   \]
   Since \( f(1) = 1 \), we have:
   \[
   f(2) = 1
   \]

4. Set \( x = 1 \) and \( y = \sqrt{2} \):
   \[
   f(1^2 \cdot (\sqrt{2})^2) = f(1^4 + (\sqrt{2})^4) \implies f(2) = f(1 + 4) \implies f(2) = f(5)
   \]
   Since \( f(2) = 1 \), we have:
   \[
   f(5) = 1
   \]

5. Set \( x = \sqrt[4]{2} \) and \( y = \sqrt[4]{2} \):
   \[
   f((\sqrt[4]{2})^2 \cdot (\sqrt[4]{2})^2) = f((\sqrt[4]{2})^4 + (\sqrt[4]{2})^4) \implies f(2) = f(2)
   \]
   This is consistent with our previous results.

6. Set \( x = \sqrt[4]{3} \) and \( y = \sqrt[4]{3} \):
   \[
   f((\sqrt[4]{3})^2 \cdot (\sqrt[4]{3})^2) = f((\sqrt[4]{3})^4 + (\sqrt[4]{3})^4) \implies f(3) = f(6)
   \]
   Since \( f(2) = 1 \), we have:
   \[
   f(3) = f(6)
   \]

7. By setting different values for \( x \) and \( y \), we can see that the function \( f \) must be constant for all positive real numbers. This is because the functional equation \( f(x^2 y^2) = f(x^4 + y^4) \) forces \( f \) to take the same value for different arguments.

8. Since \( f(1) = 1 \), the only possible function is \( f(x) = 1 \) for all \( x \in \mathbb{R}^+ \).

Conclusion:
The number of such functions is exactly one.

The final answer is \( \boxed{1} \)