For $x = y = 0$ we deduce $f^2(0) \mid 2f(0)$, whence $f(0) \in \{0, 1, 2\}$.

If $f(0) = 0$, for $y = 0$ we have $0 \mid f(x)$, $\forall x \in \mathbb{N}$, so $f(x) = 0$, $\forall x \in \mathbb{N}$, which contradicts the fact that $f$ is strictly increasing.

If $f(0) = 1$, for $y = 0$, we have $f(x) \mid (2x + 1) + f(x)$, $\forall x \in \mathbb{N}$, thus $f(x) \mid 2x + 1$, $\forall x \in \mathbb{N}$.
Assuming that $f(k) = 2k + 1$, for some natural $k$, we have $f(k + 1) \mid 2k + 3$, and since $f(k + 1) > f(k) = 2k + 1$ we get $f(k + 1) = 2k + 3$, that is $f(x) = 2x + 1$, $\forall x \in \mathbb{N}$.

If $f(0) = 2$, for $y = 0$, we have $2f(x) \mid 2(2x + 1) + f(x)$, $\forall x \in \mathbb{N}$, thus $\frac{2(2x+1)+f(x)}{2f(x)} = \frac{2x+1}{f(x)} + \frac{1}{2} \in \mathbb{N}$, $\forall x \in \mathbb{N}$.
Assuming that $f(k) = 4k + 2$, for some natural $k$, we have $\frac{2k+3}{f(k+1)} + \frac{1}{2} \in \mathbb{N}$ and since $f(k + 1) > f(k) = 4k + 2$ we get $0 < \frac{2k+3}{f(k+1)} + \frac{1}{2} < \frac{2k+3}{4k+2} + \frac{1}{2} \le 2$. So $\frac{2k+3}{f(k+1)} + \frac{1}{2} = 1$, thus $f(k+1) = 4k + 6$, that is $f(x) = 4x + 2$, $\forall x \in \mathbb{N}$.

Both functions fulfill the property in the statement.