Put $x = 0$ in the given equation
$$f(x^{2015} + f(y)^{2015}) = f(x)^{2015} + y^{2015}. \qquad (1)$$
Then, we have
$$f(f(y))^{2015} = f(0)^{2015} + y^{2015}. \qquad (2)$$
It implies that $f$ is a bijective function. By putting $f(x)$ to $x$ in (1), we have
$$f(f(x))^{2015} + f(y)^{2015} = f(f(x))^{2015} + y^{2015} \qquad (3)$$
and by putting $f(y)$ to $y$ in (1),
$$f(x^{2015} + f(f(y))^{2015}) = f(x)^{2015} + f(y)^{2015}. \qquad (4)$$
In (3), we swipe $x$ and $y$ first and then take $f$ on both sides, then we have
$$f(f(f(x))^{2015} + f(y)^{2015})) = f(x^{2015} + f(f(y))^{2015}) = f(x)^{2015} + f(y)^{2015}.$$
Since $f$ is bijective, we have
$$f(f(x)) = x. \tag{5}$$
By (2), (3), and (5), we have
$$(f(f(x))^{2015} + f(y)^{2015}) = x^{2015} + y^{2015} = f(f(x)^{2015}) + f(f(y)^{2015}) - 2f(0)^{2015}$$
and, since $f$ is bijective, $f(x + y) = f(x) + f(y) - 2f(0)^{2015}$. Here, by putting $x = y = 0$, we have $f(0) = 2f(0)^{2015}$. Hence $f(0)$ is $0$ or $\pm(\frac{1}{2})^{\frac{1}{2014}}$. By putting $f(y)$ to $y$ in (2), we get $f(y^{2015}) = f(0)^{2015} + f(y)^{2015}$.
Consider an equation $x^{2015} - x + f(0)^{2015}$. It must have three different real roots $f(-1)$, $f(0)$ and $f(1)$. However, if $f(0) = \pm(\frac{1}{2})^{\frac{1}{2014}}$, this equation does not have three different real roots. Therefore, $f(0) = 0$ and $f(1)$ is either $1$ or $-1$.
Now we have $f(x + y) = f(x) + f(y)$ and $f(y^{2015}) = f(y)^{2015}$, and $f(1)$ is either $1$ or $-1$. We will prove that only $f(x) = x$ and $f(x) = -x$ satisfy these three conditions. If $f(1) = -1$, then we may consider $g(x) = -f(x)$. It allows us to consider only the case where $f(1) = 1$. By the general Cauchy method, we have $f(x) = x$ for all $x \in \mathbb{Q}$.
Now we need to extend this function to real numbers. To do that, we will prove that $f(x^2) = f(x)^2$.
Let $\binom{2015}{k}f(x^k) - f(x)^k = x_{2015-k}$ for $k = 0, 1, 2, \dots, 2015$. Then, for any integer $m$, by calculating $f((x+m)^{2015}) = (f(x)+f(m))^{2015}$, we have
$$\sum_{i=0}^{2015} m^i x_i = 0.$$
Now we think of it as a system of infinite number of equations.
Claim. The system has a unique solution $(x_0, x_1, \dots, x_{2015}) = (0, 0, \dots, 0)$.
*Proof.* Suppose there is another solution $(y_0, y_1, \dots, y_{2015}) = (0, 0, \dots, 0)$. Let $s$ be the maximum index with $y_s \neq 0$. By multiplying $\frac{1}{y_s}$, we may assume that $y_s = 1$. Set $m$ as a number bigger than $|y_0| + |y_1| + \dots + |y_{2015}|$. Then,
$$0 = \sum_{i=0}^{2015} m^i y_i = \sum_{i=0}^{s} m^i y_i > m^s y_s - \sum_{i=0}^{s-1} m^i |y_i| \geq m^s - m^{s-1} \sum |y_i| > 0$$
which yields a contradiction. Therefore it has a unique solution $(0, 0, \dots, 0)$. $\square$

The above claim says that $f(x^k) = f(x)^k$ for $k = 0, 1, 2, \dots, 2015$. Especially, $f(x^2) = f(x)^2 \ge 0$. Now, for every $x > y$, we have $f(x) - f(y) = f(x - y) = f(\sqrt{x - y}^2) \ge 0$, so $f$ is increasing. Therefore, $f(x) = x$, and the candidates of the original problem are $f(x) = x$ and $f(x) = -x$. Indeed, one can check that they are real solutions of the problem by substituting.