If $x$ and $y$ are positive integers, and $x^4+y^4=4721$, find all possible values of $x+y$
1. We start with the equation \( x^4 + y^4 = 4721 \) and note that \( x \) and \( y \) are positive integers.
2. To find possible values of \( x \) and \( y \), we observe that one of \( x^4 \) or \( y^4 \) must be greater than half of 4721, which is 2360.5. This is because if both \( x^4 \) and \( y^4 \) were less than or equal to 2360.5, their sum would be at most \( 2 \times 2360.5 = 4721 \), which is not possible since they must sum to exactly 4721.
3. We now look for perfect fourth powers between 2360.5 and 4721. The fourth powers of integers in this range are:
   \[
   7^4 = 2401 \quad \text{and} \quad 8^4 = 4096
   \]
4. We first check \( x^4 = 8^4 = 4096 \):
   \[
   y^4 = 4721 - 4096 = 625
   \]
   Since \( 625 = 5^4 \), we have \( y = 5 \). Thus, one solution is \( (x, y) = (8, 5) \).
5. Next, we check \( x^4 = 7^4 = 2401 \):
   \[
   y^4 = 4721 - 2401 = 2320
   \]
   Since 2320 is not a perfect fourth power, there are no integer solutions for \( y \) in this case.
6. Therefore, the only valid solution is \( (x, y) = (8, 5) \) or \( (x, y) = (5, 8) \).

Conclusion:
\[
x + y = 8 + 5 = 13
\]

The final answer is \(\boxed{13}\).