Suppose there exist integers $x$ and $y$ such that $x^4 + x^3 + x^2 + x = y^6 + 61$.

Let us consider the equation modulo $7$.

First, note that $y^6 \equiv 0$ or $1 \pmod{7}$ for any integer $y$, since by Fermat's Little Theorem, $y^6 \equiv 1$ if $y$ is not divisible by $7$, and $0$ otherwise.

So $y^6 \equiv 0$ or $1 \pmod{7}$.

Now, $61 \equiv 5 \pmod{7}$, so $y^6 + 61 \equiv 5$ or $6 \pmod{7}$.

Therefore, $x^4 + x^3 + x^2 + x \equiv 5$ or $6 \pmod{7}$.

Let us compute $x^4 + x^3 + x^2 + x \pmod{7}$ for all $x = 0, 1, 2, 3, 4, 5, 6$:

- $x = 0$: $0^4 + 0^3 + 0^2 + 0 = 0$
- $x = 1$: $1 + 1 + 1 + 1 = 4$
- $x = 2$: $16 + 8 + 4 + 2 = 30 \equiv 2 \pmod{7}$
- $x = 3$: $81 + 27 + 9 + 3 = 120 \equiv 1 \pmod{7}$
- $x = 4$: $256 + 64 + 16 + 4 = 340 \equiv 4 \pmod{7}$
- $x = 5$: $625 + 125 + 25 + 5 = 780 \equiv 3 \pmod{7}$
- $x = 6$: $1296 + 216 + 36 + 6 = 1554 \equiv 1 \pmod{7}$

So the possible residues are $0, 1, 2, 3, 4$ modulo $7$.

But $y^6 + 61 \equiv 5$ or $6 \pmod{7}$, which are not among the possible residues for $x^4 + x^3 + x^2 + x$.

Therefore, there are no integer solutions to the equation.