Solution:
Let $a = x^{2} + 3x + 1$ and let $b = x^{2} - x - 6$. The only way to have $a^{b} = 1$ is if $a = \pm 1$ or $b = 0$.

- If $b = 0$, then we solve the quadratic $x^{2} - x - 6 = 0$ which has solutions $x = -2, 3$ (we would also have to check that $a \neq 0$ in this case)

- If $a = 1$, then we solve the quadratic $x^{2} + 3x + 1 = 1$ which has solutions $x = 0, -3$.

- If $a = -1$, then we solve the quadratic $x^{2} + 3x + 1 = -1$ which has solutions $x = -1, -2$ (we also have to check that $b$ is an even integer in this case)

Therefore there are a total of 5 candidate solutions: $x = -3, -2, -1, 0, 3$.