Denote $q_1 = \frac{1+\sqrt{5}}{2}$, $q_2 = \frac{1-\sqrt{5}}{2}$, and then $q_1 + q_2 = 1$, $q_1q_2 = -1$. Thus,
$$a_n = \frac{1}{\sqrt{5}}(q_1^n - q_2^n), \quad n = 1, 2, \dots$$
Hence $a_1 = 1, a_2 = 1$. Also note that $q_i + 1 = q_i^2$ ($i = 1, 2$), and there is
$$\begin{aligned} a_{n+1} + a_n &= \frac{1}{\sqrt{5}}(q_1^{n+1} - q_2^{n+1}) + \frac{1}{\sqrt{5}}(q_1^n - q_2^n) \\ &= \frac{1}{\sqrt{5}}(q_1^n(q_1+1) - q_2^n(q_2+1)) \\ &= \frac{1}{\sqrt{5}}(q_1^{n+2} - q_2^{n+2}), \end{aligned}$$
namely,
$$a_{n+2} = a_{n+1} + a_n, \quad n = 1, 2, \dots$$
It is easy to know that each term of the sequence $\{a_n\}$ is a positive integer.

It is easy to calculate that $q_1^4 + q_2^4 = 7$, and hence
$$\begin{aligned} a_{2n+3}a_{2n-1} - 1 &= \frac{1}{\sqrt{5}}(q_1^{2n+3} - q_2^{2n+3}) \cdot \frac{1}{\sqrt{5}}(q_1^{2n-1} - q_2^{2n-1}) - 1 \\ &= \frac{1}{5}(q_1^{4n+2} + q_2^{4n+2} - (q_1q_2)^{2n-1}q_1^4 - (q_1q_2)^{2n-1}q_2^4) - 1 \end{aligned}$$
$$\begin{align*} &= \frac{1}{5}(q_1^{4n+2} + q_2^{4n+2} + q_1^4 + q_2^4) - 1 \\ &= \frac{1}{5}(q_1^{4n+2} + q_2^{4n+2} + 7) - 1 \\ &= \frac{1}{5}(q_1^{4n+2} + q_2^{4n+2} + 2) \\ &= \left[ \frac{1}{\sqrt{5}}(q_1^{2n+1} + q_2^{2n+1}) \right]^2 \\ &= a_{2n+1}^2. \end{align*}$$
Therefore, for any positive integer $n$, $a_{2n+3}a_{2n-1} - 1$ is a perfect square.

Hence, $a_{m+4}a_m - 1$ are perfect squares for all positive odd numbers $m$.