Is this proof rigorous?

Is this proof rigorous?

"There is no rational number whose square is $\displaystyle \frac{m}{n}$,
where $\displaystyle \frac{m}{n}$ is a positive fraction in lowest terms,
unless $m$ and $n$ are perfect squares. For suppose, if possible, that
$\displaystyle \frac{p^2}{q^2} = \displaystyle \frac{m}{n}$, $p$ having no
factor in common with $q$ and $m$ no factor in common with n. Then $np^2 =
mq^2$. Every factor of $q^2$ must divide $np^2$ and, as $p$ and $q$ have
no common factor, every factor of $q$ must divide $n$. Hence $n=rq^2$
where $r$ is an integer. But this involves $m = rp^2$; and as $m$ and $n$
have no common factor, $r$ must be unity. Thus $m = p^2$, $n = q^2$, as
was to be proved. In particular, it follows by taking $m = 2$, $n = l$
that $2$ cannot be the square of a rational number."
A part of this proof strikes me as odd:
"Hence $n=rq^2$ where $r$ is an integer. But this involves $m = rp^2$; and
as $m$ and $n$ have no common factor, $r$ must be unity."
The use of $r$ as a factor for both $n$ and $m$ requires a presupposition
that $r$ is unity. Without this knowledge, the writer would need to assume
two distinct variables, instead of just $r$. The proof seemed logically
continuous prior to this bit, so I'm looking for second opinions.
I'm working through this book (G. H. Hardy's 'A Course of Pure
Mathematics') without guidance or consultation (excepting the fine folk on
the internet), and little more than a broken, North American public school
level education. The very 'general' mode of explanation Hardy uses here is
leading me to spend an awful lot of time second-guessing myself.