Consider the semidirect product G' = \langle X^i,Y \rangle \rtimes_{\varphi} \langle Z \rangle where $i$ ranges over $\mathbb{Z}[1/2]$, powers of $X$ multiply in the obvious way and they commute with $Y$ (which has infinite order). The element $Z$ also has infinite order, and the action of $Z$ is $X^i Y^j \to X^{2i} Y^j$ (here $i \in \mathbb{Z}[1/2], j \in \mathbb{Z})$. Then $X, Y, Z$ satisfy the given relations (with the map $x \mapsto X$, etc., of course), and by construction, all $X^i Y^j Z^k \ne 1$ for $(i,j,k) \ne (0,0,0)$, hence $x^i y^j z^k \ne 1$.
Edit: As Jack Schmidt points out in the comments, the original construction with $i \in \mathbb{Z}$ didn't work because the action of $Z$ is not an automorphism (only even powers of $X$ would be in the image).
Edit 2: As Jack Schmidt points out in the comment to Manolito's answer, this can be fixed by letting $i$ range over $\mathbb{Z}[1/2]$ (rationals whose denominator is a power of 2) instead of just $\mathbb{Z}$. The map G \to G' is surjective because $x^z$ has image $X^{1/2}$, $x^{z^k}$ has image $x^{1/2^k}$, etc., so G' is a quotient of $G$.
Edit 3: In fact, we can prove that G \cong G'. It is sufficient to show that every element of $G$ has a normal form $(x^a)^{z^b} z^c y^d$ where $a$ is an odd integer and $b, c, d$ are arbitrary integers. These elements are distinct because the image of the above element in G' is $X^{a/2^b} Z^c Y^d$, and the decomposition of an element of $\mathbb{Z}[1/2]$ into the form $a/2^b$, where $a$ is odd, is unique.
Take any word in $x,y,z$ and their inverses. First of all, since $y$ commutes with everything, we can just pull all the $y$'s out to the right. From now on I'll assume we just have $x$'s and $z$'s.
To avoid a lot of double superscripts, I'm going to write $x^{[a]}$ for $x^{z^a}$. Notice that the original relation $zx = x^2z$ can be written as $x^{[-1]} = x^2$ and also that we have the law $(x^{[a]})^{[b]} = x^{[a+b]}.$ The above identities continue to hold if $x$ is replaced by any integer power of $x$.
So we have a word in $x$'s and $z$'s (and their inverses). By repeatedly applying the identity $z^{-a} x = x^{[a]} z^{-a}$ we can pull the $z$'s out to the right as well, and we are just left with a product of elements of the form $x^{[a]}$. We must prove that such a product has the form $(x^b)^{[a]}$.
First of all, we have $(x^2)^{[a]} = (x^{[-1]})^{[a]} = x^{[a-1]}$ and the above is still valid if $x$ is replaced by any integer power of $x$, hence by iterating $(x^{2^k})^{[a]} = x^{[a-k]}$ for integers $k \ge 0$. Now the general product $x^{[a]} x^{[b]} = x^{[a]} x^{[a-(a-b)]} = x^{[a]} (x^{2^{a-b}})^{[a]} = (x^{1+2^{a-b}})^{[a]}$ valid for $a-b \ge 0$, has the desired form. If $a < b$, we rewrite $[a] = [b-(b-a)]$ instead, and do a similar calculation. The even more general product $(x^m)^{[a]} (x^n)^{[b]}$ is similar.
So elements of $G$ have the normal form $(x^a)^{[b]} y^c z^d$, and we just need to show that we can make $a$ odd. This follows from the identity $(x^{2a})^{[b]} = ((x^a)^2)^{[b]} = ((x^a)^{[-1]})^{[b]} = (x^a)^{[b-1]}$ which we can repeatedly apply to remove all the powers of 2 from $a$.