Assume that $xyzt \neq 0$.
If three of the numbers $x$, $y$, $z$, $t$ have a common divisor $d$, the remaining number is also a multiple of $d$. The division of all the four numbers by $d$ preserves the validity of our equations. Since $x$, $y$, $z$, $t$ are not zero, we can divide them by their greatest common divisor and henceforth assume that they are coprime.

The numbers $x$ and $y$ can not be both odd (since $xy$ is even) or both even (since in this case $zt$ is also even and at least three of the numbers $x$, $y$, $z$, $t$ are even). Thus $x$ and $y$ and therefore $z$ and $t$ have different parity.

Denote $s = x^2 + y^2$, $n = zt$. Without loss of generality, $n > 0$.
The condition on $x$, $y$, $z$, $t$ means that $s - 4n$, $s - 2n$, $s + 2n$, $s + 4n$ are perfect squares. We will prove that this is not possible for any odd $s$ and positive $n$. First, we find a statement equivalent to these four numbers being squares. Their product $(s^2 - 4n^2)(s^2 - 16n^2) = (s^2 - 10n^2)^2 - 36n^4$ is a square of some odd $T$. The numbers $T$, $6n^2$, $s^2 - 10n^2$ form a primitive Pythagorean triple, hence there exist coprime $u$, $v$ of different parity such that $2uv = 6n^2$ and $u^2 + v^2 = s^2 - 10n^2$. Since $u$ and $v$ are coprime, it follows from $uv = 3n^2$ that one of these numbers is a square and another is a triple square. Without loss of generality, assume that $u = 3k^2$ and $v = l^2$; $k$ and $l$ are coprime. Multiplying $u^2 + v^2 = s^2 - 10n^2$ by $3$ we get $3s^2 - 30n^2 = 3s^2 - 10uv = 3u^2 + 3v^2$, that is, $3s^2 = 3u^2 + 10uv + 3v^2 = (3u + v)(3v + u)$, where $3u + v$ and $3v + u$ are coprime (they have different parity, and their common divisor must also divide $8u = 3(3u + v) - (3v + u)$ and, similarly, $8v$). It follows that one of the numbers $3u + v$ and $3v + u$ is a square and another is a triple square. Since $3$ does not divide $3u + v$, the square is $3u + v = 9k^2 + l^2$, and the triple square is $u + 3v = 3(k^2 + l^2)$. We have proved that the original equations imply the existence of coprime $k$ and $l$ such that $9k^2 + l^2$ and $k^2 + l^2$ are perfect squares.

It remains to prove that there are no such $k$, $l$. If $k^2 + l^2$ and $9k^2 + l^2$ are perfect squares, the pairs $(k, l)$ and $(3k, l)$ are pairs of legs in primitive Pythagorean triples. Unfortunately, we do not know which of the numbers $k$ and $l$ is even, so two cases are possible. But let's prove the following lemma first.

Lemma. Let $a$, $b$, $c$, $d$ be pairwise coprime positive integers such that $c^2(a^2 + d^2) = b^2(9a^2 + d^2)$. Then, $a^2 + d^2$ and $9a^2 + d^2$ are perfect squares.

Proof. Since $\text{GCD}(b, c) = 1$, then there exists such positive integer $f$ that
$$a^2 + d^2 = b^2 f, \quad 9a^2 + d^2 = c^2 f.$$
$$f = \text{GCD}(a^2 + d^2, 9a^2 + d^2) \implies f \mid 8a^2, \quad f \mid 8d^2.$$
$$\text{GCD}(a, d) = 1 \implies f \mid 8.$$
If $a$, $d$ are of different parity, then $a^2 + d^2$ is odd, and therefore $f = 1$. Let both $a$ and $d$ be odd (they can't be simultaneously even). Then
$$a^2 + d^2 \equiv 2 \pmod{8} \implies f \mid 2.$$
Let $f = 2$.
$$9a^2 + d^2 = c^2 f = 2c^2 \implies d^2 \equiv 2c^2 \pmod{3} \implies 3 \mid c, \quad 3 \mid d$$
— a contradiction, since $\text{GCD}(c, d) = 1$. Thus, $f = 1$, $a^2 + d^2 = b^2$, $9a^2 + d^2 = c^2$. The lemma is proven.

Case 1. $k$ is even. Then $k = 2u_1v_1$, $3k = 2u_2v_2$, $l = u_1^2 - v_1^2 = u_2^2 - v_2^2$, where $(u_1, v_1)$ and $(u_2, v_2)$ are two pairs of coprime numbers with different parity. Since $u_2v_2 = 3u_1v_1$, then there exist pairwise coprime positive integers $a$, $b$, $c$, $d$ such that $u_2 = 3ab$, $v_2 = cd$, $u_1 = ac$, $v_1 = bd$ (the case when $u_2$ is not divisible by $3$ can be considered similarly). Substituting this in the expression for $l$ we obtain $a^2c^2 + c^2d^2 = 9a^2d^2 + d^2c^2$, or $c^2(a^2 + d^2) = d^2(9a^2 + c^2)$. According to the lemma, $a^2 + d^2$ and $9a^2 + c^2$ are squares, and, obviously, $a < k$, $d < l$.

Case 2. $l$ is even. Then $k = u_1^2 - v_1^2$, $3k = u_2^2 - v_2^2$, $l = 2u_1v_1 = 2u_2v_2$. Since $(u_2 - v_2)(u_2 + v_2) = 3(u_1 - v_1)(u_1 + v_1)$, then we have $u_1 - v_1 = ab$, $u_1 + v_1 = cd$, $u_2 - v_2 = ac$, $u_2 + v_2 = 3bd$ for some pairwise coprime positive integers $a$, $b$, $c$, $d$ (the case when $u_2 + v_2$ is not divisible by $3$ can be considered similarly). Expressing $u_1$, $v_1$, $u_2$, $v_2$ in terms of $a$, $b$, $c$, $d$ and putting the result in the formula for $l$, we get $(ab+cd)(cd-ab) = (3bd+ac)(3bd-ac)$, that is, $c^2(a^2+d^2) = b^2(a^2+9d^2)$. Hence, by the lemma, we obtain that $a^2+d^2$ and $a^2+9d^2$ are perfect squares, and $a<l$, $d<k$.

In both cases, for each pair $(k, l)$ such that $k^2 + l^2$ and $9k^2 + l^2$ are perfect squares we constructed a pair of smaller numbers with the same property. It follows that such numbers do not exist. Thus, the original system does not admit a solution in non-zero integers.