Proof: Given $a_{3}=48$.
For all $n$, $a_{n+1}^{2}=a_{n} a_{n+2}+1$.
$$\begin{array}{l} \text { Also, } \frac{a_{n-1}+a_{n+1}}{a_{n}}=\frac{a_{n-1} a_{n+1}+a_{n+1}^{2}}{a_{n} a_{n+1}} \\ =\frac{\left(a_{n}^{2}-1\right)+\left(a_{n} a_{n+2}+1\right)}{a_{n} a_{n+1}}=\frac{a_{n}+a_{n+2}}{a_{n+1}} \\ \Rightarrow \frac{a_{n}+a_{n+2}}{a_{n+1}}=\frac{a_{n-1}+a_{n+1}}{a_{n}}=\cdots=\frac{a_{1}+a_{3}}{a_{2}}=7 \\ \Rightarrow a_{n+2}=7 a_{n+1}-a_{n}(n=1,2, \cdots) . \end{array}$$

Given $a_{1}=1, a_{2}=7$, we can inductively show that each term of the sequence is a positive integer.
To prove that $9 a_{n} a_{n+1}+1$ is a perfect square, we observe that,
$$\begin{array}{l} 9 a_{1} a_{2}+1=64=8^{2}, \\ 9 a_{2} a_{3}+1=3025=55^{2}, \\ \ldots \ldots . \end{array}$$

Transforming this information into the structure of the sequence, we find that
$$8=a_{1}+a_{2}, 55=a_{2}+a_{3} \text {. }$$

Thus, we conjecture that for all $n \in \mathbf{N}$,
$$\begin{array}{l} 9 a_{n} a_{n+1}+1=\left(a_{n}+a_{n+1}\right)^{2} . \\ \text { Let } f(n)=9 a_{n} a_{n+1}+1-\left(a_{n}+a_{n+1}\right)^{2}(n=1,2, \cdots) . \\ \text { By } f(n)-f(n-1) \\ =9 a_{n} a_{n+1}-9 a_{n} a_{n-1}-\left(a_{n}+a_{n+1}\right)^{2}+\left(a_{n-1}+a_{n}\right)^{2} \\ =\left(a_{n+1}-a_{n-1}\right)\left(7 a_{n}-a_{n+1}-a_{n-1}\right) \stackrel{\text { (1) }}{=} 0, \end{array}$$

Therefore, $f(n)=f(n-1)=\cdots=f(1)=0$, which means equation (2) holds.