Assuming indirectly that $x y=a^{2}$ and $z t=c^{2}$ with $a, c>0$. Suppose that the number $x+y=z+t$ is odd. Then $x$ and $y$ have opposite parity, as well as $z$ and $t$. This means that both $x y$ and $z t$ are even, as well as $x y-z t=x+y$; a contradiction. Thus, $x+y$ is even, so the number $s=\frac{x+y}{2}=\frac{z+t}{2}$ is a positive integer. Next, we set $b=\frac{|x-y|}{2}, d=\frac{|z-t|}{2}$. Now the problem conditions yield
$$s^{2}=a^{2}+b^{2}=c^{2}+d^{2}$$
and
$$2 s=a^{2}-c^{2}=d^{2}-b^{2}$$
(the last equality in (2) follows from (1)). We readily get from (2) that $a, d>0$. In the sequel we will use only the relations (1) and (2), along with the fact that $a, d, s$ are positive integers, while $b$ and $c$ are nonnegative integers, at most one of which may be zero. Since both relations are symmetric with respect to the simultaneous swappings $a \leftrightarrow d$ and $b \leftrightarrow c$, we assume, without loss of generality, that $b \geqslant c$ (and hence $b>0$). Therefore, $d^{2}=2 s+b^{2}>c^{2}$, whence
$$d^{2}>\frac{c^{2}+d^{2}}{2}=\frac{s^{2}}{2}$$
On the other hand, since $d^{2}-b^{2}$ is even by (2), the numbers $b$ and $d$ have the same parity, so $00$ imply $b=c=0$, which is impossible.