Example 7 On the equation about $x, y$
$$
x^{2}+y^{2}=208(x-y)
$$

all positive integer solutions are $\qquad$
(2008, "Mathematics Weekly" Cup National Junior High School Mathematics Competition)
Since 208 is a multiple of 4, the square of an even number is divisible by 4 with a remainder of 0, and the square of an odd number is divisible by 4 with a remainder of 1, therefore, \(x\) and \(y\) are both even.
Let \(x = 2a, y = 2b\). Then \(a^2 + b^2 = 104(a - b)\).
Similarly, \(a\) and \(b\) are both even.
Let \(a = 2c, b = 2d\). Then \(c^2 + d^2 = 52(c - d)\).
Therefore, \(c\) and \(d\) are both even.
Let \(c = 2s, d = 2t\). Then \(s^2 + t^2 = 26(s - t)\).
Thus, \((s - 13)^2 + (t + 13)^2 = 2 \times 13^2\), where \(s\) and \(t\) are both even.
Hence, \((s - 13)^2 = 2 \times 13^2 - (t + 13)^2\)
\(\leq 2 \times 13^2 - 15^2 < 11^2\).
Therefore, \(|s - 13|\) could be \(1, 3, 5, 7, 9\).
Thus, \((t + 13)^2\) is
\(337, 329, 313, 289, 257\).
Hence, \((t + 13)^2\) can only be \(289\).
Thus, \(|s - 13| = 7\).