Define the Fibonacci sequence
$\displaystyle n_0\ =\ 0,\ n_1\ =\ 1$
$\displaystyle n_k\ =\ n_{k-1}+n_{k-2},\ k\geq2$
Then
$\displaystyle n_k\ =\ \frac1{\sqrt5}\left[\left(\frac{1+\sqrt5}2\right)^k-\left(\frac{1-\sqrt5}2\right)^k\right]$
as can be verified by induction (or otherwise).
Using this formula, we can check that
$\displaystyle n_{k+1}^2\ =\ \frac15\left[\left(\frac{3+\sqrt5}2\right)\left(\frac{1+\sqrt5} 2\right)^{2k} + \left(\frac{3-\sqrt5}2\right)\left(\frac{1-\sqrt5}2\right)^{2k} + 2(-1)^k\right]$
$\displaystyle n_kn_{k+1}\ =\ \frac15\left[\left(\frac{1+\sqrt5}2\right)^{2k+1} + \left(\frac{1-\sqrt5}2\right)^{2k+1} - (-1)^k\right]$
$\displaystyle n_k^2\ =\ \frac15\left[\left(\frac{1+\sqrt5}2\right)^{2k} + \left(\frac{1-\sqrt5}2\right)^{2k} - 2(-1)^k\right]$
Hence the following relation is satisfied:
$\displaystyle \boxed{n_{k+1}^2 - n_kn_{k+1} - \left[n_k^2+(-1)^k\right]\ =\ 0}$
This is a quadratic equation in $\displaystyle n_{k+1}$; its discriminant is $\displaystyle 5n_k^2+4$ when
k is even. This must be a perfect square since $\displaystyle n_{k+1}$ is an integer, and since there are infinitely many Fibonacci numbers such that
k is even, there are infinitely many integers
n such that $\displaystyle 5n^2+4$ is a perfect square.
Similarly the discriminant is $\displaystyle 5n_k^2-4$ when
k is odd; this must be a perfect square, and as there are infinitely many Fibonacci numbers such that
k is odd, there are also infinitely many integers
n such that $\displaystyle 5n^2-4$ is a perfect square.