Find all functions $f:\mathbb{Q}^{+} \to \mathbb{Q}^{+}$ such that for all $x\in \mathbb{Q}^+$: [list] [*] $f(x+1)=f(x)+1$, [*] $f(x^2)=f(x)^2$. [/list]
1. **Given Conditions:**
   We are given two functional equations for \( f: \mathbb{Q}^+ \to \mathbb{Q}^+ \):
   \[
   f(x+1) = f(x) + 1 \quad \text{for all } x \in \mathbb{Q}^+
   \]
   \[
   f(x^2) = f(x)^2 \quad \text{for all } x \in \mathbb{Q}^+
   \]

2. **Induction on \( f(x+i) = f(x) + i \):**
   We start by proving that \( f(x+i) = f(x) + i \) for all \( i \in \mathbb{N} \) using induction.
   
   - **Base Case:**
     For \( i = 1 \), we have \( f(x+1) = f(x) + 1 \), which is given.
   
   - **Inductive Step:**
     Assume \( f(x+k) = f(x) + k \) holds for some \( k \in \mathbb{N} \). Then,
     \[
     f(x + (k+1)) = f((x+k) + 1) = f(x+k) + 1 = (f(x) + k) + 1 = f(x) + (k+1)
     \]
     Thus, by induction, \( f(x+i) = f(x) + i \) holds for all \( i \in \mathbb{N} \).

3. **Analyzing \( f \left( \frac{p}{q} \right) \):**
   Let \( r = f \left( \frac{p}{q} \right) \). Consider the function evaluated at \( \frac{p}{q} + iq \):
   \[
   f \left( \frac{p}{q} + iq \right) = r + iq
   \]
   Now, consider the function evaluated at the square of this expression:
   \[
   f \left( \left( \frac{p}{q} + iq \right)^2 \right) = f \left( \frac{p^2}{q^2} + 2i \frac{p}{q} + i^2 q^2 \right)
   \]
   Using the first condition, we can write:
   \[
   f \left( \frac{p^2}{q^2} + 2i \frac{p}{q} + i^2 q^2 \right) = f \left( \frac{p^2}{q^2} \right) + 2i \frac{p}{q} + i^2 q^2
   \]
   Since \( f(x^2) = f(x)^2 \), we have:
   \[
   f \left( \frac{p^2}{q^2} \right) = f \left( \frac{p}{q} \right)^2 = r^2
   \]
   Therefore,
   \[
   f \left( \left( \frac{p}{q} + iq \right)^2 \right) = r^2 + 2i \frac{p}{q} + i^2 q^2
   \]

4. **Equating the Two Expressions:**
   On the other hand, using the second condition directly:
   \[
   f \left( \left( \frac{p}{q} + iq \right)^2 \right) = f \left( \frac{p}{q} + iq \right)^2 = (r + iq)^2 = r^2 + 2irq + i^2 q^2
   \]
   Equating the two expressions, we get:
   \[
   r^2 + 2i \frac{p}{q} + i^2 q^2 = r^2 + 2irq + i^2 q^2
   \]
   Simplifying, we find:
   \[
   2i \frac{p}{q} = 2irq
   \]
   Dividing both sides by \( 2i \), we obtain:
   \[
   \frac{p}{q} = rq
   \]
   Hence,
   \[
   r = \frac{p}{q}
   \]

5. **Conclusion:**
   The function \( f(x) = x \) satisfies both given conditions:
   \[
   f(x+1) = x+1 = f(x) + 1
   \]
   \[
   f(x^2) = x^2 = (f(x))^2
   \]
   Therefore, the only function that satisfies both conditions is \( f(x) = x \).

The final answer is \( \boxed{ f(x) = x } \)