I suppose there are infinitely many infinite sequences of integer squares, all of whose first differences are also integer squares. Here is an attempt at constructive proof.
Suppose you have the sequence up to $a_n$ and wish to extend it. Write $a_n$ as a difference of squares (it is a square): $ a_n = a^2 = (\frac{\frac{a_n}{d}+d}{2})^2 - (\frac{\frac{a_n}{d}-d}{2})^2, d \mid a_n$. Setting $a_{n+1}=(\frac{\frac{a_n}{d}+d}{2})^2$ and $a_{n+1}-a_n=(\frac{\frac{a_n}{d}-d}{2})^2$ will extend the sequence as long as it is an integer. To force it being an integer, one can insist that $a_n = 16 u^2$ with $u$ odd and take $d=4, \frac{a_n}{d}=4 u^2$ (avoiding factorization) leading to $\frac{\frac{a_n}{d}+d}{2} \equiv 0 \mod 4$ and the square again of the form $16u^2$ with $u$ odd (this follows from examining $(\frac{4+4(2x+1)^2}{2})^2 \mod 32$). So start from $a_1=16 u^2$ and extend the sequence.
After simplification, $a_{n+1}=(\frac{4+\frac{a_n}{4}}{2})^2$ and $a_1=16 u^2$, $u>1$ odd.
Starting with $a_1= 16 \cdot 5^2$ I get:
400, 2704, 115600, 208860304, 681603644851600, 7259117635546998039104028304, 823356075729834991394377343895101538985808607052531600, 10592425428769277708701964508444107521120841773208159861878488881058295592932634035770367240431209291868304