Rational power of 2

Let $r=\frac{p}{q}$ where $q$ is a positive integer, and $p$ is any integer, and let’s assume that this fraction is irreducible.
Let’s define $x=2^{r}=2^{\frac{p}{q}}$. This is an integer by assumption.

Then, we have that $x^{q}=2^{p}$. This means that the only prime factor of $x$ has to be 2. We can therefore write $x=2^{k}$ for some integer $k$; but then $x^{q}=(2^{k})^{q}=2^{kq}=2^{p}$, so $p=kq$; but this is impossible, unless $q=1$, because $\frac{p}{q}$ was supposed to be irreducible. So $r=p$ is an integer.