I am aware that the quaternion group can be presented as $\langle i,j,k \mid i^2=j^2=k^2=ijk\rangle$. When I was first trying to write a presentation for the quaternion group, I considered the following: $$\langle x,y \mid x^4=y^4=1\space,yx=xy^3\rangle$$ I know that the group formed by the above presentation has order of at most $16$. If it has order less than $16$, it must have order $8$ as we can set $x=i$ and $y=j$ to get the quaternion group. Moreover, if it has order $8$, we need to show $x^2=y^2$.
How do we prove that the above presentation forms a group of order $16$? If that is untrue, how do we show $x^2=y^2$?
This group has order $16$. Take a semidirect product $C_4\rtimes C_4$, with the generator of the $C_4$ on the right acting by inversion on the $C_4$ on the left. This is a group of order $16$ that satisfies your presentation.