Q:

Clarification on proof that orthogonal matrices have $-1$ and $1$ as eigenvalues

All orthogonal matrices have an eigenvalue of $1$ and $-1$. The proof provided is as follows.
Let an orthogonal matrix Q have eigenvalues $\lambda$.
Then $\mathbf{Qx}=\lambda\mathbf{x}$ and since Q is orthogonal,
$\mathbf{Ix}=\mathbf{Q^TQx}=\mathbf{Q^T}(\lambda \mathbf{x}) = \lambda(\mathbf{Q^Tx})=\lambda^2\mathbf{x}$.
Why is $\mathbf{Q^Tx} = \lambda \mathbf{x}$?

A:

It might be possible that you've mixed up different results?
The determinant of an orthogonal matrix is always $\pm 1$:
$${\bf M}^{\top}{\bf M} = {\bf I}$$
$$\det\left({\bf M}^{\top}{\bf M}\right) = \det\left({\bf I}\right)$$
$$\det({\bf M})\det({\bf M}^{\top}) = 1$$
$$\left[\det({\bf M})\right]^2 = 1$$
$$\det({\bf M}) = \pm 1$$

