 Solution

Yes: the displayed presentation already has p-deficiency exactly one for $p=2$. The words $x,y,z,xy,yz,zx$ are not proper powers in the free group. For the length-two words this follows directly from their distinct consecutive letters in a cyclically reduced word. Thus each relator's $2$-root exponent is precisely the exponent of $2$ in its displayed power. The relator weights, in the given order, areTheir sum is $2$, givingIn particular the infinitude criterion proves that this group is infinite, although the question only asks for the existence of the presentation and prime.