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:
1.2. Elementary proof. The following elementary proof of theorem 1 is outlined by Dickson in , 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:
show that we have:
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
IV. The conclusion in the last step is incompatible with the positivity of $a,b,c,d$.