Find the largest value of $a^b$ such that the positive integers $a,b>1$ satisfy $$a^bb^a+a^b+b^a=5329$$
1. We start with the given equation:
   \[
   a^b b^a + a^b + b^a = 5329
   \]

2. Add 1 to both sides of the equation:
   \[
   a^b b^a + a^b + b^a + 1 = 5330
   \]

3. Notice that the left-hand side can be factored:
   \[
   (a^b + 1)(b^a + 1) = 5330
   \]

4. We need to find pairs of positive integers \(a, b > 1\) such that the product of \((a^b + 1)\) and \((b^a + 1)\) equals 5330. First, we factorize 5330:
   \[
   5330 = 2 \times 5 \times 533 = 2 \times 5 \times 13 \times 41
   \]

5. We look for pairs of factors of 5330 that are of the form \(a^b + 1\) and \(b^a + 1\). The factors of 5330 are:
   \[
   1, 2, 5, 10, 13, 26, 41, 65, 82, 106, 205, 410, 533, 1066, 2665, 5330
   \]

6. We need to check which pairs of these factors can be written as \(a^b + 1\) and \(b^a + 1\). We start with the pair \(65\) and \(82\):
   \[
   65 = 4^3 + 1 \quad \text{and} \quad 82 = 3^4 + 1
   \]

7. Verify if \(a = 4\) and \(b = 3\) satisfy the original equation:
   \[
   a^b = 4^3 = 64 \quad \text{and} \quad b^a = 3^4 = 81
   \]
   \[
   a^b b^a = 64 \times 81 = 5184
   \]
   \[
   a^b + b^a = 64 + 81 = 145
   \]
   \[
   a^b b^a + a^b + b^a = 5184 + 64 + 81 = 5329
   \]

8. Since the equation holds true, the largest value of \(a^b\) is:
   \[
   4^3 = 64
   \]

The final answer is \(\boxed{64}\).