The hint I was given was to simply prove that $y=xz$ is irrational given that $x$ is nonzero, $x$ is rational and $z$ is irrational. Here's how I did it:
Claim: $y=xz$ is irrational.
Proof: Assume $x\neq0$, $x$ is rational and $z$ is irrational.
By contradiction assume that $y=xz$ is rational. This means $y$ can be expressed as $m/n$, $m$ and $n$ being integers; $y$ can be expressed similarly as $p/q$, $p$ and $q$ being integers. By substitution, we have that $$ p/q=mz/n$$ and $$z=pn/qm, qm \neq 0.$$ Since $pn$ and $qm$ are integers $z$ has to be rational.
In addition to this it seems like there's a part 2 as follows:
Proof: Given an interval $(x,y)$ we will choose a positive irrational number, $z$, say. By density of the rationals there is a rational $p$ in the interval $(x/z, y/z)$ s.t. $$ x/z <p< y/z.$$ From this we see that $pz$ is irrational since it is the product of a rational and irrational number.
Is the $pz$ the $xz$ that we proved is irrational in the first proof? So ideally when presenting a full proof like this, should we do part 2 then part 1?