elementary number theory - The contradiction method used to prove that the square root of a prime is irrational

The contradiction method given in certain books to prove that sqare root of a prime is irrational also shows that sqare root of $4$ is irrational, so how is it acceptable?
e.g. Suppose $\sqrt{4}$ is rational,
$$\begin{align} \sqrt{4} &=p/q \qquad\text{where pand q are coprimes} \\ 4 &=p^2/q^2\\ 4q^2&=p^2 \tag{1} \\ 4&\mid p^2\\ 4&\mid p\\ \text {let }p&=4m \qquad\text{for some natural no. m} \\ p^2&=16m^2\\ 4q^2&=16m^2 \qquad\text{(from (1) )}\\ q^2&=4m^2\\ 4& \mid q^2\\ 4&\mid q \end{align}$$

but this contradicts our assumption that $p$ and $q$ are coprime since they have a common factor $p$. Hence $\sqrt{4}$ is not rational. But we know that it is a rational. Why?