Find all increasing functions $f$ from the nonnegative integers to the integers satisfying $f(2)=7$ and \[ f(mn) = f(m) + f(n) + f(m)f(n) \] for all nonnegative integers $m$ and $n$.
1. **Given Conditions and Initial Setup:**
   We are given that \( f \) is an increasing function from the nonnegative integers to the integers, satisfying:
   \[
   f(2) = 7
   \]
   and
   \[
   f(mn) = f(m) + f(n) + f(m)f(n)
   \]
   for all nonnegative integers \( m \) and \( n \).

2. **Proposing a Candidate Function:**
   Let's propose \( f(x) = x^3 - 1 \) and verify if it satisfies the given conditions.
   - For \( f(2) \):
     \[
     f(2) = 2^3 - 1 = 7
     \]
     which matches the given condition.
   - For the functional equation:
     \[
     f(mn) = (mn)^3 - 1
     \]
     and
     \[
     f(m) + f(n) + f(m)f(n) = (m^3 - 1) + (n^3 - 1) + (m^3 - 1)(n^3 - 1)
     \]
     Simplifying the right-hand side:
     \[
     (m^3 - 1) + (n^3 - 1) + (m^3 - 1)(n^3 - 1) = m^3 + n^3 - 2 + (m^3 - 1)(n^3 - 1)
     \]
     \[
     = m^3 + n^3 - 2 + m^3 n^3 - m^3 - n^3 + 1
     \]
     \[
     = m^3 n^3 - 1
     \]
     which matches \( f(mn) = (mn)^3 - 1 \).

3. **Transforming the Functional Equation:**
   Define \( g(x) = f(x) + 1 \). Then the functional equation becomes:
   \[
   g(mn) = f(mn) + 1 = (f(m) + 1)(f(n) + 1) = g(m)g(n)
   \]
   This implies that \( g \) is a multiplicative function.

4. **Analyzing \( g \):**
   - For \( g(0) \):
     \[
     g(0) = g(0)^2 \implies g(0) \in \{0, 1\}
     \]
     If \( g(0) = 1 \), then \( g(0) = g(0)g(n) \) for all \( n \), implying \( g \equiv 1 \), which contradicts \( g(2) = 8 \). Thus, \( g(0) = 0 \).
   - For \( g(1) \):
     \[
     g(1) = g(1)^2 \implies g(1) \in \{0, 1\}
     \]
     If \( g(1) = 0 \), then \( g(n) = g(1)g(n) = 0 \) for all \( n \), which is absurd. Thus, \( g(1) = 1 \).

5. **Determining \( g(x) \):**
   - We know \( g(2) = f(2) + 1 = 8 \).
   - We need to show \( g(x) = x^3 \) for \( x \geq 3 \).

6. **Using Sequences to Prove \( g(x) = x^3 \):**
   - Consider sequences \( a_n, b_n \) such that \( \frac{a_n}{b_n} \to \log_2(x) \) from below and above.
   - For \( a_n < \log_2(x)b_n \):
     \[
     2^{a_n} < x^{b_n} \implies g(2^{a_n}) < g(x^{b_n}) \implies 8^{a_n} < g(x)^{b_n}
     \]
     \[
     8^{a_n / b_n} < g(x) \implies 8^{\log_2(x)} = x^3 \leq g(x)
     \]
   - For \( c_n > \log_2(x)d_n \):
     \[
     2^{c_n} > x^{d_n} \implies g(2^{c_n}) > g(x^{d_n}) \implies 8^{c_n} > g(x)^{d_n}
     \]
     \[
     8^{c_n / d_n} > g(x) \implies x^3 = 8^{\log_2(x)} \geq g(x)
     \]
   - Combining both, we get \( x^3 \geq g(x) \geq x^3 \), hence \( g(x) = x^3 \).

7. **Conclusion:**
   Since \( g(x) = x^3 \), we have \( f(x) = g(x) - 1 = x^3 - 1 \).

The final answer is \( \boxed{ f(x) = x^3 - 1 } \)