There's a famous proof that $\sqrt{2}$ is irrational by assuming $\sqrt{2}=p/q$ for relatively prime $p$ and $q$ and then proving that this leads to $p$ and $q$ being both even which contradicts with them being coprime.
Now there's something I noticed that may make another easier proof:
Assume for the sake of contradiction that there exists positive integers $p$ and $q$ such that $\gcd(p,q)=1 \implies \gcd(p^2,q^2)=1$, and
$$\sqrt{2}=\frac{p}{q} \implies 2q^2=p^2 \implies q^2|p^2$$
Now the last result contradicts with $p^2$ and $q^2$ being coprime, except if $q^2=1 \implies q= 1 \implies p^2=2$, but $2$ is not a perfect square so there's no such $p$
Is this proof correct?
Edit (adding a comment): this can be generalized that for any positive integer $n>1$, the $n$th root of a non-$n$th-powered integer is irrational.
For example: let $k\ne m^n$ be a positive integer for any positive integer $m$, we have $\sqrt[n]{k}$ is irrational, because assume for the sake of contradiction (for coprime positive integers $p,q$, we have $\gcd(p,q)=1 \implies \gcd(p^n,q^n)=1$ because $p$ and $q$ have different prime factors and thus so is $p^n$ and $q^n$, considering the fundamental theorem of arithmetic)
$$\sqrt[n]{k}=p/q \implies kq^n=p^n \implies q^n|p^n$$
this, similarly, contradicts with $\gcd(p^n,q^n)=1$ and implies $q^n=1$ and so $q=1$ so $k=p^n$, a contradiction.
Best Answer
Yes, it's valid, provided you address @razivo's point. Indeed, any prime factor dividing both $p^2$ and $q^2$ divides both $p$ and $q$.
In case you're interested, there are several other well-known proofs. (It looks like that list omits the proof by the rational root theorem.)
I'll leave it to others to say when "two proofs" are different enough to be different proofs.