Squares in arithmetic progression
Here are a few proofs: 1, and the somewhat bizarre 3. I'd previously linked to Kiming's exposition to prove this result, but the link has been removed. This is the proof described in lhf's answer --- and I think of this as a very elementary approach.
Unfortunately, there are no cases where you have nontrivial arithmetic progressions of higher powers. This is a string of proofs. Carmichael himself covered this for n = 3 and 4, about a hundred years ago. But it wasn't completed until Ribet wrote a paper on it in the 90s. His paper can be found here. The statement is equivalent to when we let $\alpha = 1$. Funny enough, he happens to have sent out a notice on scimath with a little humor, which can still be found here.
A quick Google search found this paper: On 4 Squares in Arithmetic Progression by Ian Kiming. It contains a sketch of an elementary proof at the end and cites Dickson's History of the theory of numbers. It is reproduced below.
There do not exist $4$ rational squares in arithmetic progression:
Theorem 1. Suppose that $\alpha$, $\beta$, $\gamma$ and $\delta$ are rational numbers such that: $$\beta^2-\alpha^2=\gamma^2-\beta^2=\delta^2-\gamma^2.$$ Then $\pm\alpha=\pm\beta=\pm\gamma=\pm\delta$.
1.2. Elementary proof. The following elementary proof of theorem 1 is outlined by Dickson in [9], p. 440. As an exercise, fill in the details of the following steps.
I. To prove theorem 1 assume that we are given rational numbers $\alpha$, $\beta$, $\gamma$ and $\delta$ satisfying the hypothesis of the theorem. We may assume that they are relatively prime, non-negative integers. The theorem then states that $\alpha=\beta=\gamma=\delta$. Assume that this is not true. Then we see that $\alpha^2<\beta^2<\gamma^2<\delta^2$.
Show that $\gcd(\alpha,\beta)=\gcd(\beta,\gamma)=\gcd(\gamma,\delta)=1$ and that $\alpha$, $\beta$, $\gamma$, $\delta$ are all odd. Conclude that $\gcd(\beta+\alpha,\beta-\alpha)=\gcd(\gamma+\beta,\gamma-\beta)=\gcd(\delta+\gamma,\delta-\gamma)=2$.
II. Defining positive integers $a,b,c,d$ by: $$2a:=\gcd(\beta+\alpha,\gamma+\beta),\\ 2b:=\gcd(\beta+\alpha,\gamma-\beta),\\ 2c:=\gcd(\beta-\alpha,\gamma+\beta),\\ 2d:=\gcd(\beta-\alpha,\gamma-\beta),$$ show that we have: $$\beta+\alpha=2ab,\\ \beta-\alpha=2cd,\\ \gamma+\beta=2ac,\\ \gamma-\beta=2bd,\\ \delta+\gamma=2bc,\\ \delta-\gamma=2ad.$$
III. Then $(a+d)b=(a-d)c$ and $(c+d)a=(c-d)b$, and we have the number $\gcd(a+d,a-d)$ and $\gcd(c+d,c-d)$ are both either $1$ or $2$. We can then conclude that $$a+d=mc,\\a-d=mb,\\c+d=nb,\\c-d=na,$$ with $m,n\in\{1,2\}$.
IV. The conclusion in the last step is incompatible with the positivity of $a,b,c,d$.
My favourite proof of this is Van der Poorten's — it uses descent, as Fermat almost certainly would have.