Proof that $\frac{1 + \sqrt{5}}{2}$ is irrational.
Another completely different approach: One can easily see (by squaring the golden ratio) that it satisfies the quadratic equation $x^2-x-1=0$. Using the Rational Root Theorem we concude that it must be irrational.
Another approach:
Continued fractions are finite for rationals. Now, try to find the continued fraction representation for $\frac{1+\sqrt{5}}{2}\big($you may use the fact that it satisfies the equation $x^2-x-1=0\big)$, it will be $[1;1 ,1,1,\cdots] $, which has infinitely many $1$. Conidering the contrapositive statement of the statement above, if a number has infinitely many terms in it's cont. fraction representation then it is not rational. Hence, $\frac{1+\sqrt{5}}{2} $ is not rational or irrational.
Another one:
Let, $\frac{1+\sqrt{5}}{2}=\frac{p}{q},p,q\in\mathbb{Z}$, then $\sqrt{5}=\frac{2p-q}{q}$. Now, we can prove that $\sqrt{5}$ is irrational using this argument:
$\sqrt{5}$ satisfies the monic polynomial $x^2-5=0$, hence, if it is an rational algebraic number it need to be an integer. But, $2<\sqrt{5}<3$ (as, $4<5<9$). Hence, $\sqrt{5}$ is irrational.
Now come back to the equation $\sqrt{5}=\frac{2p-q}{q}$. As, $p,q\in\mathbb{Z}$, $\frac{2p-q}{q}\in\mathbb{Q}$, as, both $2p-q,q\in\mathbb{Z}$. But, we have shown that $\sqrt{5}$ is irrational. Hence, contradiction!
Yes, this is basically how you prove that $\sqrt n$ is irrational. In general the argument is that, if $n$ is not a square it has at least prime factor $p$ of odd exponent in the factorization of $n$, but then if $a^2=b^2 n$, and so the exponent of $p$ is odd on the right but even on the left, contradiction. This is just a generalization of your argument.
About the way you proved that if $p|n^2$ then $p|n$ for prime $p$ via FTA, not only is this not an inelegant method, I think it's basically how anyone who knows number theory would do it, and it's probably the method that most clearly explains "why" the theorem is true. Whatever simpler method you know for the case $p=2$, it's probably either just this method in disguise, or it involves a property of the number $2$ that really doesn't generalize to other primes.