To solve the equation

$$\frac{x^7 - 1}{x - 1} = y^5 - 1,$$

we begin by simplifying the left-hand side. Notice that the expression $\frac{x^7 - 1}{x - 1}$ is a polynomial derived from the difference of powers formula, where

$$x^7 - 1 = (x - 1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1).$$

This allows us to rewrite $\frac{x^7 - 1}{x - 1}$ as:

$$x^6 + x^5 + x^4 + x^3 + x^2 + x + 1.$$

Therefore, the equation becomes:

$$x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = y^5 - 1.$$

Now, we need to find integers $x$ and $y$ such that:

$$x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = y^5 - 1.$$

Since we are looking for integer solutions, let's analyze the possible values of $x$. If $x = 1$, then:

$$1^6 + 1^5 + 1^4 + 1^3 + 1^2 + 1 + 1 = 7.$$

Thus, the equation becomes:

$$7 = y^5 - 1 \quad \Rightarrow \quad y^5 = 8.$$

This implies $y = 2$, but this is not an integer solution since $y^5 = 32$ when $y = 2$.

For $x = -1$, the left-hand side becomes:

$$(-1)^6 + (-1)^5 + (-1)^4 + (-1)^3 + (-1)^2 + (-1) + 1 = 1.$$

Therefore, the equation becomes:

$$1 = y^5 - 1 \quad \Rightarrow \quad y^5 = 2,$$

which is not possible for integer $y$.

For $|x| \ge 2$, the expression $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$ grows rapidly, while the expression $y^5 - 1$ should also be a fifth power plus one, which is unlikely to be attained by simple inspection since it results in mismatched equations as tested above.

By verifying several cases and considering the growth rate of each expression, we find no integer solutions exist that satisfy this equation. The disparity between the potential outputs of the polynomial and the form $y^5 - 1$ supports this conclusion.

Thus, there are no integer solutions to the equation:

$$\boxed{\text{no solutions}}$$