Let $x, y$ be positive integers such that
$$
x^4=(x-1)\left(y^3-23\right)-1 .
$$
Find the maximum possible value of $x+y$.
1. Start by rearranging the given equation:
   \[
   x^4 = (x-1)(y^3 - 23) - 1
   \]
   Adding 1 to both sides, we get:
   \[
   x^4 + 1 = (x-1)(y^3 - 23)
   \]

2. Since \(x\) and \(y\) are positive integers, \(x-1\) must be a divisor of \(x^4 + 1\). We also know that \(x-1\) is a divisor of \(x^4 - 1\). Therefore, \(x-1\) must divide the difference:
   \[
   (x^4 + 1) - (x^4 - 1) = 2
   \]
   Hence, \(x-1\) must be a divisor of 2. The possible values for \(x-1\) are 1 and 2 (since \(x-1\) must be positive).

3. Consider the case \(x-1 = 1\):
   \[
   x-1 = 1 \implies x = 2
   \]
   Substituting \(x = 2\) into the original equation:
   \[
   2^4 + 1 = (2-1)(y^3 - 23)
   \]
   \[
   16 + 1 = y^3 - 23
   \]
   \[
   17 = y^3 - 23
   \]
   \[
   y^3 = 40
   \]
   Since 40 is not a perfect cube, there are no integer solutions for \(y\) in this case.

4. Consider the case \(x-1 = 2\):
   \[
   x-1 = 2 \implies x = 3
   \]
   Substituting \(x = 3\) into the original equation:
   \[
   3^4 + 1 = (3-1)(y^3 - 23)
   \]
   \[
   81 + 1 = 2(y^3 - 23)
   \]
   \[
   82 = 2(y^3 - 23)
   \]
   \[
   41 = y^3 - 23
   \]
   \[
   y^3 = 64
   \]
   \[
   y = 4
   \]
   Therefore, \(x = 3\) and \(y = 4\).

5. The maximum possible value of \(x + y\) is:
   \[
   x + y = 3 + 4 = 7
   \]

The final answer is \(\boxed{7}\)