Instead of doing matrix calculation you can also map those qubits to the Bloch sphere. A Bloch sphere is a unit 2-sphere, with antipodal points corresponding to a pair of mutually orthogonal state vectors.
So if you can show that the 2 points are antipodal on a Bloch sphere, then you have proven that they are orthogonal.
The nice thing is that your qubits are already expressed in a way that can easily be mapped on the bloch sphere (x, y, n and m correspond to the angles on the Bloch sphere).
So giving the angles it is easy to determine if they are antipodal as in that case the corresponding angles must differ with $\pi + n.2\pi$ where n is a positive or negative natural number (note that in case $\theta =0 + n\pi$ is a special case as in that case we should ignore angle $\psi$).
In the question it is already stated that the angles m and n differ with $\pi$ so we only need to look at the angles x and y and so we can conclude that only under the following condition will the above 2 qubits be orthogonal
- $y-x = { \pi + n (2 \pi)}$ where $n$ is a positive or negative natural number [1]
As $x+y= \pi$ (see question) and [1] : then only the following x and y combinations will give orthogonal qubits.
- $2y = \pi + \pi + n (2 \pi)$ which can be simplified as $y = n\pi$ where n is a positive or negative natural number and in that case $x=\pi - n\pi$ (e.g. when $y=0$ en $x=\pi$ we get 2 orthogonal qubits).