Let $(x, y)$ be any solution of the given system. Since $\sqrt{\sqrt{x} + 2}$ is obviously positive, we have $y > 2$ by the first equation. Similarly, the second equation implies $x > 2$.

Now we prove that the numbers $x$ and $y$ must be equal. We will use the observation that the function square root is increasing. If $x > y$, then
$$\sqrt{\sqrt{x} + 2} > \sqrt{\sqrt{y} + 2},$$
i.e. $y - 2 > x - 2$, $y > x$, and that is a contradiction. The case $x < y$ is eliminated similarly. The equality of $x = y$ is thus proved.

Let us therefore deal with the (only possible) case $x = y$ next. The original system of two equations is then obviously reduced to a single equation
$$\sqrt{\sqrt{x} + 2} = x - 2. \qquad (1)$$
After substituting $s = \sqrt{x}$, when $x = s^2$, the equation (1) becomes $\sqrt{s+2} = s^2-2$, while obviously $s > \sqrt{2}$. For each such $s$, we square the equality to obtain $s+2 = (s^2-2)^2$, which we rewrite in the form $s^4 - 4s^2 - s + 2 = 0$. Note that this equation has a root $s = 2$. This is confirmed by the decomposition
$$s^4 - 4s^2 - s + 2 = s^2(s^2 - 4) - (s - 2) = s^2(s - 2)(s + 2) - (s - 2) = (s - 2)(s^3 + 2s^2 - 1),$$
by which we now show that $s = 2$ is the only root of the derived equation, that satisfies our condition $s > \sqrt{2}$. Indeed, for every $s > \sqrt{2}$, $s^3 + 2s^2 - 1 > 0$ (this is even true for $s \ge 1$). The only satisfactory value of $s = 2$ corresponds to the only solution $x = s^2 = 4$ of the equation (1), and therefore to the only solution $x = y = 4$ of the given problem.

Conclusion. The given system of equations has a unique solution $(x, y) = (4, 4)$.

We again use the observation that both numbers $x$ and $y$ are greater than 2, and introduce the function $f: (2, \infty) \to (2, \infty)$, $f(t) = \sqrt{t} + 2$ for every $t > 2$. The equations from the problem, rewritten in the form
$$\begin{aligned} \sqrt{\sqrt{x} + 2} + 2 &= y, \\ \sqrt{\sqrt{y} + 2} + 2 &= x \end{aligned}$$
can then be written as a system of equations using the function $f$
$$\begin{aligned} f(f(x)) &= y, \\ f(f(y)) &= x. \end{aligned}$$
We see that its solutions are just pairs of the form $(x, y) = (x, f(f(x)))$, where the number $x$ satisfies the relation $f(f(f(x))) = x$. This is certainly satisfied in the case where $f(x) = x$. We show that the equality holds only in this case.

If $f(x) < x$ we have a quadruple of inequalities
$$f\left(f\left(f\left(f(x)\right)\right)\right) < f\left(f\left(f(x)\right)\right) < f(f(x)) < f(x) < x$$
where the last inequality is obvious and every previous inequality is the consequence of the immediately following inequality and the fact that $f$ is increasing. Similarly in the case of $f(x) > x$ we have
$$f\left(f\left(f(f(x))\right)\right) > f\left(f(f(x))\right) > f(f(x)) > f(x) > x.$$
Thus, we have proved equivalence of $f\left(f\left(f(f(x))\right)\right) = x$ and $f(x) = x$.

It remains to solve the equation $f(x) = x$ with unknown $x > 2$, which is easy:
$$f(x) = x \Leftrightarrow \sqrt{x} + 2 = x \Leftrightarrow 0 = (\sqrt{x} - 2)(\sqrt{x} + 1) \Leftrightarrow \sqrt{x} = 2 \Leftrightarrow x = 4.$$
We arrive at the same conclusion as in the first solution.