proof verification - Show that $sqrt{q}$ is irrational

Let q be a positive integer such that $q \geq 2$ and such that for any

integers a and b, if $q|ab$, then $q|a$ or $q|b$. Show that $\sqrt{q}$
is irrational.

Proof;

Let assume $\sqrt{q}$ is a rational number, where $n \neq 0$ and $\gcd (m,n)=1$, meaning $\sqrt{q} = \frac{m}{n} \Rightarrow q=\frac{m^2}{n^2} $

Since $n^2 \nmid m^2$, $q|m^2 \Rightarrow q|m$, so $m=qt$ where $t\in \mathbb{Z}$

By substitute $m=qt$ in the equation $qn^2 = m^2$, we get $n^2=qt^2$.

Since tells us that $q|n^2$ and $t^2|n^2$, it contradicts with the assumption $\gcd (m,n)=1$; therefore, $\sqrt{q}$ is irrational.

I get this proof with the assistant of the course, but is there any flaw or mistake? What are the other methods for proving this statement, can you at least give one different method? And how can I improve this proof?