"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
Other apporoach can be $N \times 2^{N+1} + 1 = k^2$. Hence,$k$ is in the form of $2m+1$ and we will have: $N \times 2^{N+1} + 1 = 4m^2 +4m + 1$. Then, $N \times 2^{N+1} = 4 \times m \times (m+1) \Rightarrow N \times 2^{N-1} = m \times (m+1)$. Now, we have a simple proof that $gcd(m, m+1) = 1$, and one of them is odd and te other is even. Now we have the following case:
$N$ is odd
$N$ is even
We can factorize $N = 2^{\alpha_0} q_1^{\alpha_1}q_2^{\alpha_2}\cdots q_i^{\alpha_i}$. However, like the previous cases, if we decompose the left side to two even and odd numbers, we will reach an inequality such that one side of it is greater than $2^{N-1}$, and the other side is less than $N-1$. Therefore, in that case, we will find $N \leq 3$ as well. The only even number in this range is 2, but it does not satisfy the condition. Hence, there is no answer when $N$ is even.