$(x, y) = (1,1)$ is the unique real solution of this system of equations.
Clearly, $(x, y) = (1,1)$ is a real solution of this system of equations. We now prove that there is no other real solution.
Suppose otherwise, that $(x, y) = (x_1, y_1) \neq (1,1)$ is another real solution of the system of equations
$$\begin{cases} x + 2^x a_n + 2^y a_{n+1} = 1 + 2a_{n+2}, \\ y + 2^x a_{n+1} + 2^y a_{n+2} = 1 + 2a_{n+3}, \end{cases}$$
For convenience, let $(x_2, y_2) = (1,1)$. Substituting these two solutions into the system of equations and subtracting, we obtain respectively
$$(x_1 - x_2) + (2^{x_1} - 2^{x_2})a_n + (2^{y_1} - 2^{y_2})a_{n+1} = 0, \quad (1)$$
$$(y_1 - y_2) + (2^{x_1} - 2^{x_2})a_{n+1} + (2^{y_1} - 2^{y_2})a_{n+2} = 0, \quad (2)$$
Multiplying (1) by $(2^{x_1} - 2^{x_2})$ and (2) by $(2^{y_1} - 2^{y_2})$, then adding the two together and simplifying, we obtain
$$(x_1 - x_2)(2^{x_1} - 2^{x_2}) + (y_1 - y_2)(2^{y_1} - 2^{y_2}) \\ + (2^{x_1} - 2^{x_2})^2 a_n + 2(2^{x_1} - 2^{x_2})(2^{y_1} - 2^{y_2})a_{n+1} + (2^{y_1} - 2^{y_2})^2 a_{n+2} = 0. \quad (3)$$
We now prove that when $(x_1, y_1) \neq (x_2, y_2)$, (3) does not hold, thereby leading to a contradiction.
Note that $f(x) = 2^x$ is an increasing function, hence we obtain
$$(x_1 - x_2)(2^{x_1} - 2^{x_2}) \geq 0 \quad \text{and} \quad (y_1 - y_2)(2^{y_1} - 2^{y_2}) \geq 0.$$
Equality holds in both of the above inequalities simultaneously if and only if $(x_1, y_1) = (x_2, y_2)$.
Therefore, when $(x_1, y_1) \neq (x_2, y_2)$, we obtain the following strict inequality:
$$(x_1 - x_2)(2^{x_1} - 2^{x_2}) + (y_1 - y_2)(2^{y_1} - 2^{y_2}) > 0. \quad (4)$$
On the other hand, using mathematical induction it is easy to prove that when $n$ is odd, the following identity holds:
$$a_{n+1}^2 - a_n a_{n+2} = -1.$$
(This is a basic property of the Fibonacci sequence, and the proof is omitted here.)
And noting that $a_n > 0, (x_1, y_1) \neq (x_2, y_2)$, we thus obtain
$$(2^{x_1} - 2^{x_2})^2 a_n + 2(2^{x_1} - 2^{x_2})(2^{y_1} - 2^{y_2})a_{n+1} + (2^{y_1} - 2^{y_2})^2 a_{n+2} \quad (5) \\ = \frac{1}{a_n} \left[ (2^{x_1} - 2^{x_2})a_n + (2^{y_1} - 2^{y_2})a_{n+1} \right]^2 + \frac{1}{a_n} (2^{y_1} - 2^{y_2})^2 > 0.$$

(4) & (5) give
$$(x_1 - x_2)(2^{x_1} - 2^{x_2}) + (y_1 - y_2)(2^{y_1} - 2^{y_2}) \\ + (2^{x_1} - 2^{x_2})^2 a_n + 2(2^{x_1} - 2^{x_2})(2^{y_1} - 2^{y_2})a_{n+1} + (2^{y_1} - 2^{y_2})^2 a_{n+2} > 0,$$
which contradicts (3). In conclusion, this system of equations has exactly one real solution $(x,y) = (1,1)$.