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.

It is too late to answer but I asked a similar question .so I am starting from your steps.You are using the assumption that square of $N$ (and $M$) are $\textit{three digits}$.
Let $N^{2}=(10a+b)^{2}=100x+10y+z$ where $x,y,z$ are non negative integers and  $x,z \neq 0$.
Since $M^{2}$ is the reverse of $N^{2}$ then $M^{2}= 100z+10y+x$
If you are using one more assumption that $a>b$ then $N^{2}-M^{2}$ will lead to $a^{2}-b^{2} \leq 8$ because $x,z \neq 0$.
Now by substituting values foe $b$ and solving we get values for $a$.Then you can find result stated in starting of your problem is not even valid for all two digit numbers less then $33$ . Actually it holds good for only 21 and 31.If we use digits need not be distinct then result is very simple.
But I am not sure about two digit numbers whose square is a four digit number
