Let $N$ and $M$ be two digit numbers. Then the digits of $M^2$ are those of $N^2$, but reversed. 
Let $N$ be a two digit number and let $M$ be the number formed from $M$ by reversing $N$'s digits. The digits of $M^2$ are precisely those of $N^2$, but reversed.

$Proof$:

Since $N$ is a two digit number, we can write $N = 10a + b$ where $a$ and $b$ are the digits of $N$. Since $M$ is formed from $N$ by reversing digits, $M = 10b + a$.
$N^2 = (10a + b)^2 = 100a^2 + 20ab + b^2 $. The digits of $N^2$ are $a^2, 2ab, b^2$.
$M^2 = (10b + a)^2 = 100b^2 + 20ab + a^2$. The digits of $M^2$ are $b^2, 2ab, a^2$, exactly the reverse of $N^2$.

This proposition is false. Let $N$ be $15$. That means the proof above is not correct, but I can't see where exactly.

however there exists an infinite sequence of such numbers: we have
$12^2=144$
$102^2=10404$
$1002^2=1004004$
$10002^2=100040004$
$\dots\ \dots\ \dots$
and the same with digits reversed:
$21^2=441$
$201^2=40401$
$2001^2=4004001$
$20001^2=400040001$
$\dots\ \dots\ \dots$
