I think I have discovered a proof for the irrationality of $\sqrt{2}$, but I'm not sure

That's a correct proof, assuming that you've proved your "necessary condition". The proof of that necessary condition, however, may be just as complex as the more traditional proofs that $\sqrt{2}$ is irrational.

The "only one" part is easy, for if both small numbers are divisible by 4, then the large number is too, and it's not "in simplest form". The "one" part is less obvious (at least to me, at this moment, before I've had that first shot of caffeine).

This is an addition to John Hughes answer that proves the necessary condition. Since $a^2 + b^2 = c^2$, if $c$ is odd we have that at least one of $a, b$ must be even, and thus at least one of $a^2$, $b^2$ divisible by 4. That part is easy.

On the other hand if $c^2$ is even both $a$ and $b$ can be odd, and without ruling this possibility out for a primitive $(a, b, c)$ the proof is not complete. It should be obvious that if $a$, $b$ are both even the triplet wasn't primitive.

There is a way to complete your proof. Consider the equation $\bmod 4$, with $a, b$ both odd:

$$a^2 + b^2 \equiv c^2 \mod 4$$

Regardless if $a \equiv 1$ or $a\equiv 3 \mod 4$, since $3^2 \equiv 1 \mod 4$ we have:

$$1 + 1 \equiv c^2 \mod 4$$

But $c$ is even, thus

$$2 \equiv 0 \mod 4$$

which shows there are no solutions, thus $a, b$ can't both be odd with even $c$.