This is a proof that I’ve seen, not one I’ve come up with; I don’t have a reference, but I imagine it’s relatively well-known. It’s not the usual proof, but I like this one because it doesn’t rely on an infinite regress.

The following works for 2, but also (with some work) for any integer that isn’t a perfect square, or (with some more) for any rational number that isn’t one square divided by another.

Proposition: $\sqrt{2}$ is irrational.

Proof: Suppose for the sake of contradiction that $\sqrt{2} = \frac{a}{b}$, where $a$ is a non-negative integer and $b$ a positive integer.

Then $a^2 = 2b^2$.

In the prime factorisation of $a^2$ and of $b^2$, each prime appears an even number of times. In particular, 2 appears an even number of times in $a^2$ and an odd number of times in $2b^2$. By the fundamental theorem of arithmetic, the two sides cannot be equal, hence our supposition cannot be true $\blacksquare$