Solution: There are no positive integer solutions.
Let $xy = a^2$ and $zt = c^2$.
If $x+y=z+t$ is odd, then $xy$ and $zt$ are both even, so $xy-zt = x+y = z+t$
is also even, a contradiction. Let $s = \frac{x+y}{2}$; from the above it follows that $s$ is an integer. Let $b = \frac{|x-y|}{2}$, $d = \frac{|z-t|}{2}$,
then from the original conditions we obtain:
$$s^2 = a^2 + b^2 = c^2 + d^2$$
and
$$2s = a^2 - c^2 = d^2 - b^2.$$
Since in the above two equations, $a, d$ and $b, c$ are symmetric, we only need to prove that the above two equations have no solutions under the condition that $a, s, d$ are positive integers,
$b, c$ are non-negative integers, and not both zero. By symmetry, we may assume $b \ge$
$c$. We have $d^2 = 2s + b^2 > c^2$, so
$$d^2 > \frac{c^2 + d^2}{2} = \frac{s^2}{2}$$
Also
$$2s = d^2 - b^2 \geq d^2 - (d-2)^2 = 4(d-1)$$
so we have
$$\frac{s}{\sqrt{2}} < d \le \frac{s}{2} + 1$$
from which we know that $s < 2\sqrt{2} + 2 < 5$. Because when $1 \le s \le 4$, $s^2$ can only be decomposed in the form $s^2 + 0^2$,
checking shows there is no solution.