Here is my attempt at proving $$\mathbb{Z}_4 \times \mathbb{Z}_2= \langle x, y \mid x^4=y^2=1, x^y=x \rangle$$
Let $F$ be a free group upon $\{x,y\}$. Define a homomorphism $\theta: F \rightarrow \mathbb{Z}_4 \times \mathbb{Z}_2$ as follows: $$x\theta = (1,1),~ y\theta = (2,0)$$ Then by Von Dyck's theorem, there exists an epimorphism $$\phi: \langle x, y \mid x^4=y^2=1, x^y=x \rangle \rightarrow \mathbb{Z}_4 \times \mathbb{Z}_2$$
Now I just need to show both groups have the same cardinality in order to conclude $\phi$ is an isomorphism.
How can I do this?
Since you have the relations $x^y = x$ and $y^2 = 1$, you know that $x$ and $y$ commute, and so everything in the presented group may be brought to the form $\{x^i y^j \mid 1 \leq i \leq 4, 1 \leq j \leq 2 \}$, and so the cardinality of the presented group is at most 8. Since $\phi$ is a surjection, you know the cardinality is at least $|\mathbb{Z}_4 \times \mathbb{Z}_2| = 8$.