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

EDIT: About rational squares whose all k-th differences are square.

Set $a=\frac{p}{q}, p^2-q^2=u^2$

Let $a_n=a^{2n}$. All kth differences are of the form $ (a^2-1)^m a^{2s}$ and $(a^2-1)$ is a square by the choice of $p,q$. Numerical experiments support this for $a=\frac{5}{4}$ for the first 1000 terms and all differences.

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

EDIT: About rational squares whose all k-th differences are square.

Set $a=\frac{p}{q}, p^2-q^2=u^2$

Let $a_n=a^{2n}$. All kth differences are of the form $ (a^2-1)^k a^{2s}$ and $(a^2-1)$ is a square by the choice of $p,q$.

EDIT2 A possible approach to get square first and second differences of rational squares is to use the above construction and try to insert a term at the beginning (to get rid of the geometric progression). Solving symbolically leads to rational points on an elliptic curve and if it is of positive rank this will give infinitely many new solutions. For $a=\frac{5}{4}$ the first term of $(\frac{1595}{1924})^2$ was found. This might be a counter example to Gjergji Zaimi's conjecture about similar triangles (in the comments).

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

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

EDIT: About rational squares whose all k-th differences are square.

Set $a=\frac{p}{q}, p^2-q^2=u^2$

Let $a_n=a^{2n}$. All kth differences are of the form $ (a^2-1)^m a^{2s}$ and $(a^2-1)$ is a square by the choice of $p,q$. Numerical experiments support this for $a=\frac{5}{4}$ for the first 1000 terms and all differences.