Find all integer solutions to the equation $y^k = x^2 + x$, where $k$ is a natural number greater than $1$.
1. Consider the equation \( y^k = x^2 + x \), where \( k \) is a natural number greater than 1.
2. Rewrite the equation as \( y^k = x(x + 1) \).
3. Note that \( x \) and \( x + 1 \) are consecutive integers, hence they are coprime, i.e., \( \gcd(x, x+1) = 1 \).
4. Since \( y^k \) is a \( k \)-th power, both \( x \) and \( x + 1 \) must be \( k \)-th powers of some integers. Let \( x = a^k \) and \( x + 1 = b^k \) for some integers \( a \) and \( b \).
5. This implies \( a^k + 1 = b^k \).
6. For \( k > 1 \), consider the equation \( a^k + 1 = b^k \). This can be rewritten as \( b^k - a^k = 1 \).
7. For \( k = 2 \), the equation becomes \( b^2 - a^2 = 1 \), which factors as \( (b - a)(b + a) = 1 \).
8. The only integer solutions to \( (b - a)(b + a) = 1 \) are \( b - a = 1 \) and \( b + a = 1 \), leading to \( b = 1 \) and \( a = 0 \), or \( b = 0 \) and \( a = -1 \).
9. Therefore, the solutions for \( k = 2 \) are \( (a, b) = (0, 1) \) and \( (a, b) = (-1, 0) \), corresponding to \( x = 0 \) and \( x = -1 \).
10. For \( k > 2 \), consider the equation \( b^k - a^k = 1 \). By Mihăilescu's theorem (formerly Catalan's conjecture), the only solution in natural numbers for \( a^k - b^k = 1 \) with \( k > 2 \) is \( (a, b, k) = (3, 2, 2) \), which does not fit our case since \( k > 2 \).
11. Therefore, for \( k > 2 \), there are no integer solutions.

Conclusion:
There are no integer solutions for \( k > 2 \). For \( k = 2 \), the only integer solutions are \( x = 0 \) and \( x = -1 \).

The final answer is \( \boxed{ x = 0 } \) and \( x = -1 \).