For questions like this, the conventional physics notation is easier to work with than the QIT gate notation. Define $\vec \sigma = (\sigma_1,\sigma_2,\sigma_3)$ to represent the three Pauli matrices
$$\sigma_1 = X = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \;\;\; \sigma_2 = Y = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}, \;\;\;
\sigma_3 = Z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}.$$
The Pauli matrices form a basis for the Lie algebra $\mathfrak{su}_2$, and the corresponding Lie group elements, $U \in SU(2)$, are given by the exponential map
$$U = e^{i\, \vec \phi \, \cdot \vec \sigma}, \;\;\; \vec \phi \in \mathbb{R}^3.$$
Define the vector $\vec \phi$ by a magnitude $\alpha$ and unit vector $\hat \phi = (\phi_1,\phi_2,\phi_3)$ such that $\vec \phi = \alpha \hat \phi$. Simple multiplication shows that $(\vec \phi \cdot \vec \sigma)^2 = \alpha^2$. With this relationship, the Taylor expansion of $U$ works out very nicely.
$$U = \sum \limits_{i=0}^\infty \frac{i^n}{n!} \, (\vec \phi \cdot \vec \sigma)^n = \sum \limits_{j=0}^\infty \frac{(-1)^j}{(2j)!} \alpha^{2j} + i \hat \phi \cdot \vec \sigma \sum \limits_{j=0}^\infty \frac{(-1)^j}{(2j + 1)!} \, \alpha^{2j+1}$$
$$=I \, \cos \alpha + i \hat \phi \cdot \vec \sigma \sin \alpha = \begin{bmatrix} \cos \alpha + i \phi_3 \sin \alpha && (\phi_2 + i \phi_1) \sin \alpha \\ (-\phi_2 + i \phi_1) \sin \alpha && \cos \alpha - i \phi_3 \sin \alpha \end{bmatrix}.$$

With this formula it's simple to find the group element corresponding to given Lie algebra parameters. In the case of your specific question $\alpha = \tfrac{\pi}{2}$ and $\hat \phi = (\tfrac{1}{\sqrt{2}}, 0, \tfrac{1}{\sqrt{2}})$. Plugging this in gives
$$U_{x+z} = \begin{bmatrix} \frac{i}{\sqrt{2}} && \frac{i}{\sqrt{2}} \\ \frac{i}{\sqrt{2}} && -\frac{i}{\sqrt{2}} \end{bmatrix} = \frac{i(X + Z)}{\sqrt{2}}.$$
In the underlying question from qiskit, $\equiv$ is defined as equivalence modulo global phase, so, as desired, the result equals $\tfrac{X+Z}{\sqrt{2}}$ up to a global phase of $e^{i\frac{\pi}{2}}$.

The more general question of determining what other Lie algebra parameterizations share this property (apart from trivial solutions, which are given by multiples of the identity) reduces to finding solutions to the equation
$$e^{i \vec \phi \, \cdot \, \vec \sigma} = e^{i \theta} \, \hat \phi \cdot \vec \sigma \; (\text{mod} \; \theta).$$
This requires $\cos \alpha = 0$, which means vectors solving this equation will have $\alpha = \pm \tfrac{\pi}{2}$. From there it's relatively straightforward to see that solutions take the form of vectors with $\alpha = \pm \frac{\pi}{2}$ and $e^{\pm i \frac{\pi}{2} \hat \phi \, \cdot \, \vec \sigma} = \pm i \, \hat \phi \cdot \vec \sigma$.

That make sense, thank you @gls. – walid – 2019-10-23T19:12:17.603

What do you mean by: " given any direction

nwith|n|=1, denoting withσithei-thPauli matrix, you have (n⋅σ)^2=(n⋅σ)" ? – walid – 2019-10-23T19:30:54.070@glS is there any policy regarding exercise questions like this similar to the ones in say Physics SE? As I see it, you basically repeated both my and answer and the answer of ChainedSymmetry and I already felt that my hint should have been more than enough for anyone really working on this question to solve it. – Marsl – 2019-10-23T19:38:58.643

I think you were looking for $(\hat n \cdot \sigma)^2 = I$. As stated that's not accurate. It might also be worth noting that equation in the cited example is not wrong, but as stated in the question (without the definition of $\equiv$) it's ambiguous. – Jonathan Trousdale – 2019-10-23T19:44:51.780

@ChainedSymmetry you are right, of course! I'll add a remark along those lines – glS – 2019-10-23T19:56:37.893

1

@Marsl there isn't, but you might want to check out the recent discussion on meta about adding a vote-to-close reason for "not enough effort questions". Not that I think this question would fall into that category, mind you. My personal opinion is that a question should stand as long as it is a useful contribution to the site (i.e. it might help someone looking for the answer to a similar problem in the future)

– glS – 2019-10-23T20:01:13.167