Suppose we have a unit square $ABCD$. If we have a primitive Pythagorean triple $(a, b, c)$, let $CD = 1$ and hence $ED = a/b$. It also follows that $EC$ is rational of course.
In fact, using the similarity of $\Delta AFE$ and $\Delta ECD$, all of the sides will also be rational numbers except for $FC$. After some work, $FC^2 = 1^2 + \left(1 - \frac{a}{b} + \frac{a^2}{b^2} \right)^2$ $ = \frac{b^4 + (a^2 - ab + b^2)^2}{b^4}$. Thus we need $(b^2)^2 + (a^2 - ab + b^2)^2 = k^2$ for some positive integer $k$ for $FC$ to be rational:
Using Euclid's formula for Pythagorean triples, one option is that $p^2 - q^2 = b^2; \ 2pq = (a^2 - ab + b^2); \ p^2 + q^2 = k$, where integers $p > q > 0$ and $p, q$ are of opposite parity (otherwise the triple will not be primitive). Wolfram Alpha then uses the quadratic formula to tell me that $a = \frac{1}{2} \left(b \pm \sqrt{-3p^2 + 8pq + 3q^2} \right)$, which simplifies to $a = \frac{1}{2} (b \pm \sqrt{(3q - p)(3p + q)})$. This tells us that $3q > p$ or that $p/3 < q < p$.
Of course, the other option is that $p^2 - q^2 = a^2 - ab + b^2, 2pq = b^2$. This gives $a = \frac{1}{2} \left(b \pm \sqrt{2(2p+q)(p-2q)}\right)$ and hence $p > 2q$ or that $p/2 > q > 0$.
So can we find a configuration where $FC$ is rational, or where $(b^2)^2 + (a^2 - ab + b^2)^2 = k^2$ for some positive integer triple $(a, b, k)$?
It is impossible.
Plug in $c^2=a^2+b^2 \tag{1}$ to get $$(c^2-ab)^2+(b^2)^2=k^2$$ $$(c^4-2abc^2)+(a^2b^2+b^4)=k^2$$ $$c^2(a-b)^2+b^2c^2=k^2$$ $$(a-b)^2+b^2=\frac{k^2}{c^2}=m^2 \tag{2}$$ Since $a^2$ and $b^2$ could only be $0$ or $1$ mod 4, with some casework on equation (1) and (2), we could see that $a=4a_1$ and $b=4b_1$. And this would lead to an infinite descending chain.