(1) By assumption, $a_1 = 5$ and $\{a_n\}$ is strictly increasing with
$$2a_{n+1} - 7a_n = \sqrt{45a_n^2 - 36}.$$
Square both sides, and we get
$$a_{n+1}^2 - 7a_n a_{n+1} + a_n^2 + 9 = 0, \qquad \textcircled{1}$$
$$a_n^2 - 7a_{n-1}a_n + a_{n-1}^2 + 9 = 0, \qquad \textcircled{2}$$
$$\textcircled{1}-\textcircled{2}: a_{n+1} = 7a_n - a_{n-1}. \qquad \textcircled{3}$$
It follows from $a_0 = 1$, $a_1 = 5$ and ③ that $a_n$ is a positive integer for each $n \in \mathbb{N}$.

(2) From ①, we get $(a_{n+1} + a_n)^2 = 9(a_n a_{n+1} - 1)$,
$$\text{SO} \qquad a_{n+1}a_n - 1 = \left(\frac{a_{n+1} + a_n}{3}\right)^2$$
By (1), $a_n$, $a_{n+1}$ are positive integers and therefore $\frac{a_{n+1} + a_n}{3}$ is a rational number. Since $\left(\frac{a_{n+1} + a_n}{3}\right)^2 = a_{n+1}a_n - 1$ is a positive integer, so is $\frac{a_{n+1} + a_n}{3}$. Thus $a_{n+1}a_n - 1$ is the square of an integer.