Let $a = |AB|$, $b = |BC|$, $c = |CD|$, $d = |DA|$ be the side lengths of $Q$, and $f = |AC|$, $g = |BD|$ the lengths of its diagonals. It's well-known and easy to prove that
$$f^2 = \frac{ab(c^2 + d^2) + cd(a^2 + b^2)}{ab + cd},$$
and
$$g^2 = \frac{ad(b^2 + c^2) + bc(a^2 + d^2)}{ad + bc}.$$
For instance, to obtain the formula for $g$, apply the Cosine Rule in each of the triangles $ABD$ and $BCD$. Then
$$\cos A = \frac{a^2 + d^2 - g^2}{2ad}, \quad \cos C = \frac{b^2 + c^2 - g^2}{2bc}.$$
But, since $Q$ is cyclic, $\cos A = -\cos C$, whence
$$\begin{aligned} 0 &= 2abcd(\cos A + \cos C) \\ &= bc(a^2 + d^2 - g^2) + ad(b^2 + c^2 - g^2) \\ &= bc(a^2 + d^2) + ad(b^2 + c^2) - (ad + bc)g^2. \end{aligned}$$
Therefore
$$g^2 = \frac{ad(b^2 + c^2) + bc(a^2 + d^2)}{ad + bc},$$
i.e., $g^2 = \lambda(a^2 + d^2) + \mu(b^2 + c^2)$, where
$$\lambda = \frac{bc}{ad + bc}, \quad \mu = \frac{ad}{ad + bc},$$
so that $\lambda, \mu$ are positive numbers that sum to 1. Thus $g^2$ is a convex sum of $a^2 + d^2$ and $b^2 + c^2$, and so $g^2 \le \max(a^2 + d^2, b^2 + c^2) \le 2\max(a^2, d^2, b^2, c^2)$, whence
$$g \le \sqrt{2} \max(a, b, c, d).$$

Suppose $a = |AB|$ is the longest side, and the diagonal $g$ is equal to $\sqrt{2}a$. Then
$$2a^2 = \lambda(a^2 - d^2) + \mu(2a^2 - b^2 - c^2) = 0.$$
But $0 < \lambda, \mu$ and $a^2 \ge \max(b^2, c^2, d^2)$. Hence $d = b = c = a$ and $Q$ is a rhombus. Because $Q$ is cyclic, it must then be a square.