It is enough to prove that there exist $a, b \in \mathbb{N}^*$ with $(a, b) = 1$ such that:
$$\frac{9n-1}{n+7} = \frac{a^2}{b^2} \qquad (1)$$
From this relation we get:
$$n = \frac{7a^2 + b^2}{9b^2 - a^2} = \frac{7(a^2 - 9b^2) + 64b^2}{9b^2 - a^2} = -7 + \frac{64b^2}{9b^2 - a^2} \quad (2)$$
Since $(a, b) = 1$, it follows that $(a^2, b^2) = 1$ and $(9b^2 - a^2, b^2) = 1$, and hence from (2) we get that $n$ is integer, if and only if $9b^2 - a^2$ is a divisor of 64.
Since $a, b$ and $n$ are positive integers, it follows that $9b^2 - a^2 \ge 8$, and hence:
$$9b^2 - a^2 = (3b + a)(3b - a) \in \{8, 16, 32, 64\}. \quad (3)$$
Moreover, the factors $3b + a$, $3b - a$ have sum a multiple of 6 and difference a multiple of 2 and $3b + a > 3b - a$. Therefore from relation (3) the possible cases are the following:
$$\begin{aligned} (3b + a, 3b - a) &= (4, 2) \quad \eta \quad (3b + a, 3b - a) = (8, 4) \quad \eta \quad (3b + a, 3b - a) = (16, 2) \\ \Leftrightarrow (a,b) &= (1,1) \quad \eta \quad (a,b) = (2,2) \quad \eta \quad (a,b) = (7,3). \end{aligned}$$
The pair $(a,b) = (2,2)$ is rejected, because $\text{gcd}(2,2) = 2 \neq 1$, and therefore we have the values $n=1$ or $n=11$.