First, complete the square for \(g(x) = x^2 - 6x + 7\).

Rewrite \(x^2 - 6x + 7\) as \((x-3)^2 - 9 + 7\).

This simplifies to \((x-3)^2 - 2\).

Set \(y = (x-3)^2 - 2\) and solve for \(x\).

\(y + 2 = (x-3)^2\)

\(x - 3 = \pm \sqrt{y + 2}\)

Since \(x > 4\), choose the positive root: \(x = 3 + \sqrt{y + 2}\).

Thus, \(g^{-1}(x) = 3 + \sqrt{x + 2}\).

The domain of \(g^{-1}\) is \(x > -1\) because \(x + 2 > 0\).