I assume you want the quick-and-ready answer Gibbs was teaching his maritime-inclined spherical-law-of-cosines Yale students in the 1880s. With the advent of Pauli matrices for the fundamental rep of SU(2), this reduced to the standard group composition law of SU(2) , albeit the final answer is always a mess in terms of the original variables.
A rotation by angle θ around an axis $\hat{e}$ is represented by a Gibbs vector $\vec{f}=\hat{e} \tan (\theta/2)$. So, for your first rotation, $ \vec{f}=\hat{z} \tan (\theta/2)$, and for your second one, $\vec{g}=\hat{x} \tan (\phi/2)$.
The composition of the two rotations then amounts to
$$
\frac{\vec{f}+\vec{g}-\vec{f}\times\vec{g}}{1-\vec{f}\cdot \vec{g}} ~.
$$
In your specific case, the dot product in the denominator vanishes, while the cross product in the numerator is in the pure y direction, which might illustrate why the eigenvector with eigenvalue 1 of your 3×3 matrix product appeared messy.
- In any case, (Olinde Rodrigues, 1840), the axis boils down to the (un-normalized!) half-angle vector
$$
\hat{x} \sin\phi/2 \cos \theta/2+\hat{z} \cos\phi/2 \sin \theta/2 -\hat{y}\sin \phi/2 \sin\theta/2 .
$$
Nevertheless, the combined rotation angle $\gamma$ is much simpler, as you see from the Pauli-matrix WP expression, namely a degenerate spherical law of cosines (the spherical analog of the Pythagorean theorem),
$$
\cos \gamma/2 = \cos\theta/2 \cos \phi/2 ,
$$
basically admiralty stuff. Cf. this answer, and this chapter.
To sum up, your new rotation axis is, indeed, left unchanged by the succesion of your two rotations,
$$\begin{pmatrix} 1 & \\
&\cos\phi & -\sin\phi\\
&\sin \phi & \cos \phi
\end{pmatrix} \begin{pmatrix}\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta\\
& & 1
\end{pmatrix} \begin{bmatrix}\cos\theta/2 \sin\phi/2\\
-\sin\theta/2 \sin\phi/2\\
\sin\theta/2 \cos\phi/2
\end{bmatrix} =\begin{bmatrix}\cos\theta/2 \sin\phi/2\\
-\sin\theta/2 \sin\phi/2\\
\sin\theta/2 \cos\phi/2
\end{bmatrix} .$$
- Reality check: Take $\theta=\phi=\pi/2$ and watch $(1,-1,1)\mapsto (1,1,1)\mapsto(1,-1,1)$ in two steps.