Proving $\sqrt2$ is irrational

Conceptually the proof compares the parity of the $\it\color{#90f}{unique}$ powers of $\,2\,$ on each side of $\,P^2 = 2Q^2.\,$ We explain how to view it in terms of unique factorization - but only using the single prime $\,p = 2$.

A simple induction shows every natural $> 0$ can be written $\it\color{#90f}{uniquely}$ in the form $\, 2^a n\,$ for odd $\,n.\,$ Thus $\, P = 2^a p,\ Q = 2^b q\,$ for $\,p,q\,$ odd, so $\, P^2 = 2 Q^2\,$ $\Rightarrow$ $\,2^{\color{#c00}{2a}} p^2 = 2^{\color{#0a0}{1+2b}} q^2$ for $\,p^2,q^2\,$ odd, contradicting said $\it\color{#90f}{uniqueness}$, because the LHS has $\rm\color{#c00}{even}$ power of $2$ vs. $\rm\color{#0a0}{odd}$ power on RHS.

Remark $ $ The above employed unique factorization $\,2^a n\,$ is a single prime $(p=2)$ special case of the Fundamental Theorem of Arithmetic (existence and uniqueness of prime factorizations) whose proof is much easier than the general case involving arbitrary primes.

I'm sorry, but try to understand this proof, which is the traditional one.

$\sqrt{2} = \dfrac{m}{n}\ $ where $\textrm{gcd}(m,n) = 1$

$2n^2 = m^2 \Rightarrow 2 \mid m^2 \Rightarrow 2\mid m \Rightarrow m^2 = 4L^2$

$2n^2 = 4L^2 \Rightarrow 2 \mid n^2 \Rightarrow 2 \mid n \Rightarrow \textrm{gcd}(m,n) \not = 1$ which is a contradiction.

Edit: $\sqrt{2} = \frac{m}{n}$.

If $\textrm{gcd}(m,n) = d>1$ then we can write $m = dx$ and $n = dy \Rightarrow \sqrt{2} = \dfrac{dx}{dy} = \dfrac{x}{y}$.

Then $\textrm{gcd}(x,y) = 1$. You can prove this by assuming for the sake of contradiction that it is not and showing that you get a divisior of $m,n$ which is greater than $d$.

Your approach is basically correct, but far too complicated, and it leads you into some trouble. For example, you write that $\frac{f_1^2}{f_2^2}$ is odd, but it may not be an integer, and you don't say what it means for a fraction to be odd. These problems can be repaired, but I recommend taking a simpler approach altogether.

You made the most important observation: we can write an integer as the product of an odd number and a power of $2$. So $p = 2^k a$, $q=2^l b$, where $a,b$ are odd integers, and $k,l$ are nonnegative integers.

Then $p^2 = 2q^2 \iff 2^{2k}a^2 = 2^{2l+1} b^2$. Since the number of powers of $2$ dividing an integer is unique, we have $2k=2l+1$. But an even integer cannot equal an odd integer, contradiction.

This is not any different from your proof—the core idea and most of the logic is nearly identical—but it avoids a lot of troubles by not breaking into cases.