For even $\sqrt n=2m$, $n=(2m)^2=4m^2\equiv\color{blue}0\mod 4$.
For odd $\sqrt n=2m+1$, $n=(2m+1)^2=4m^2+4m+1\equiv\color{blue}1\mod 4$.
So you never achieve congruency to $\color{blue}3$, as there are no other cases.
As a byproduct, this also establishes that $4k+2$ is never a perfect square.
"Here I am stuck as to what I can do. One option is just checking all the squares from 32→99 according to the needed conditions. " Well, don't check the odd ones. And don't check $\sqrt{1000} \le n^2 \le \sqrt {1999}$ or $\sqrt{8889}\le n^2 \le \sqrt{9999}$ so that tells us to only check $46$ through $94$.
Let $n = 10a + b$ then $(10a + b)= 100a^2 + 20ab + b^2$
$b^2 = 0,4,16,36,64$. Now if $b = 4$ or $6$ then $b^2$ will cause an odd digit to be carried of the the tens place. And the tens place will be determined by $2ab$ plus the odd number added. This results in an odd number. So that is impossible.
$b^2 = 0,4,$ or $64$ and $b= 0,2,$ or $b=8$.
If $b = 0$ then we need $a^2=0,4,16,36,64$ to be a perfect square with two even digits. That can only be $a = 8$. So
So $80^2 = 6400$ is one such number.
Now we just need to check $48, 52,58, 62,68, 72,78,82,88,92$.
But $2889...3999$ all have od digits so we don't have to check $53.. 63$. Or $\sqrt {4889}...\sqrt{5999}$ or $70... 77$. Or $\sqrt{6889}...{7999}$ or $83..89$.
So we only need to check $48,52,78,82,92$
$(10a + 2)^2 = 100a^2 + 40a + 4$ which means the value carried by $4a$ whether odd or even must make $a^2$ odd of even. so the digit carried by $4a$ and an d $a$ must be the same parity.
$4\cdot 5=20$ so $2$ is even but $5$ is odd. So $(50+2)^2 = 2500 + 40\cdot 5 + 4$ so the $2$ carried by $40\cdot 5$ to $2500$ will make $2700$ i.e $2704$.
$4\cdot 8 =32$. The $3$ is odd but $8$ is even so the first two digits of $82^2$ will be $8^2 +3$ which is odd.
$4\cdot 9=36$ and $3$ is odd as is $9$ so this is good. $92^2 = 8100 + 360 + 4=8464$ with the $3$ and 81$ combining to make an even.
That just leaves $48$ and $78$ to check. It's easier to just check them then to make a carrying rule. $48^2 = 2304$. Nope. and $78^2=6084$. Good.
So $80^2 = 6400,92^2 = 8464$ and $78^2 = 6084$ are the only $3$.
Best Answer
Update: When considering the equation $a^2+b^2+c^2=d^2$ modulo $4$, you use the fact that squares are congruent to 0 or 1 modulo 4: every even number squares to be 0 modulo 4, and every odd number squares to be congruent to 1 modulo 4.
Showing $a^2,b^2,c^2$ and $d^2$ cannot all be 1 mod 4 thus only shows one of $a^2,b^2,c^2,d^2$ is divisible by $4$, or equivalently one of $a,b,c,d$ is even (i.e. divisible by $2$).
You need to argue that at least two of $a,b,c,d$ must be even. This just requires refining the analysis you have already done though!