The answer is $\alpha = 0$. In this case, the function $f(x) = x$ satisfies the statement. From now on we assume $\alpha > 0$. Note that if such a function $f$ exists, then it is strictly increasing: indeed, taking $z > y$, there is $x \in \mathbb{R}^+$ such that $z = x^{\alpha} + y$, from where we obtain:
$$f(z) = f(x^{\alpha} + y) = (f(x + y))^{\alpha} + f(y) > f(y).$$

CLAIM 1: $f$ is unbounded.
Letting $x = 1$, we obtain $f(y + 1) = (f(y + 1))^{\alpha} + f(y) \ge f(1)^{\alpha} + f(y)$. So $f(y + 1) - f(y) \ge f(1)^{\alpha}$, and we can prove (by telescopic summation) that $f(n) - f(1) \ge (n - 1)(f(1))^{\alpha}$ for all $n \in \mathbb{N}$, from which we can conclude that $f$ is unbounded.

If $\alpha = 1$, then clearly there is no such function $f$. Let us consider two cases:
Case 1: $\alpha > 1$. In this case, taking $0 < x < 1$, we have:
$$x + y > x^{\alpha} + y \Rightarrow f(x + y) > f(x^{\alpha} + y) \Rightarrow (f(x + y))^{\alpha} > (f(x^{\alpha} + y))^{\alpha}.$$
But from the original equation we know that $f(x^{\alpha} + y) > (f(x + y))^{\alpha}$, whence we conclude that $f(x^{\alpha} + y) > (f(x^{\alpha} + y))^{\alpha}$. Making $x^{\alpha} + y = z$, we get $f(z) > (f(z))^{\alpha}$, for all $z \in \mathbb{R}^{+}$ (because every positive real can be written in the form $x^{\alpha} + y$ with $0 < x < 1$). As $\alpha > 1$, we conclude that $f(z) < 1$ for all $z \in \mathbb{R}^{+}$. But this contradicts the fact that $f$ is unbounded.

Case 2: $0 < \alpha < 1$. In this case, we take $x > 1$. Then:
$$x + y > x^{\alpha} + y \Rightarrow f(x + y) > f(x^{\alpha} + y) \Rightarrow (f(x + y))^{\alpha} > (f(x^{\alpha} + y))^{\alpha}.$$

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As in the previous case, since every $z > 1$ can be written as $x^{\alpha} + y$ with $x > 1$, we obtain $f(z) > (f(z))^{\alpha}$ for all $z > 1$. In this case, as $0 < \alpha < 1$, we conclude that $f(z) > 1$ for all $z > 1$.
CLAIM 2: For all $k \in \mathbb{N}$, if $z > 1$, then $f(z) > k$.
(This implies that such a function cannot exist.)
The proof is by induction. We have already proved the base case $k = 1$. Now, suppose that $f(z) > k$, for all $z > 1$. Then, taking $y > 1$ such that $z = x^{\alpha} + y$, we obtain:
$$f(z) = f(x^{\alpha} + y) = (f(x + y))^{\alpha} + f(y) > 1 + k,$$
and the induction is complete. Therefore, the only possible value is $\alpha = 0$.