1. Define the problem and initial setup:
We need to find a bijection $f: \mathbb{N}^{+} \rightarrow \mathbb{N}^{+}$ such that there exists a positive integer $k$ (in this case, $k = 2$) and a coloring of the positive integers with $k$ colors such that for any $x \neq y$, $f(x) + y$ and $f(y) + x$ are not the same color.

2. Reformulate the condition:
Let $u = f(x) + y$ and $v = x + f(y)$. The condition $u$ and $v$ not being the same color translates to requiring $x + y$ and $f(x) + f^{-1}(y)$ to be of different colors, provided $y \neq f(x)$.

3. Initial assignments and coloring:
- Set $f(1) = 2$ and $f(3) = 1$.
- This imposes that the color of $f(1) + f^{-1}(1) = 2 + 3 = 5$ must be different from the color of $1 + 1 = 2$.
- Color $2$ white and $5$ black.

4. Define the extension step:
- Let $n$ be the first natural number where $f$ is not fully defined (i.e., either $f(n)$ or $f^{-1}(n)$ or both are not defined).
- Consider the case where $f(n)$ does not exist. Let $D$ be the subset of naturals where $f^{-1}$ is defined.
- Ensure that $f(n) + f^{-1}(x)$ and $n + x$ are of different colors for any $x \in D$.
- Denote $D' := \{n + x : x \in D\}$.

5. **Coloring and defining $f$:**
- If there exists $y \in D'$ which is not yet colored, color it arbitrarily.
- Choose $N$ big enough such that the minimal element of the set $\{N + f^{-1}(x) : x \in D\}$ is bigger than the maximal colored natural number.
- Define $f(n) := N$ and for any $x \in D$, color $f(n) + f^{-1}(x)$ in the color opposite to the color of $n + x$.

6. **Handle the case when $f^{-1}(n)$ is not defined:**
- Ensure $f^{-1}(n) + f(x)$ is of a different color compared to $n + x$.

7. **Handle the case when both $f(n)$ and $f^{-1}(n)$ are not defined:**
- First define $f(n)$ as above.
- Then apply the same step to define $f^{-1}(n)$.

8. Iterate the extension step:
- Apply the extension step consecutively to define $f$ over $\mathbb{N}^{+}$.
- If there are still uncolored naturals, color them arbitrarily.

By following these steps, we ensure that $f$ is a bijection and the coloring condition is satisfied.