"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 $3$.
Let $x$ represent the first two digits and $y$ represent the second two digits. Thus
$100x+y=x^2+y^2$
$x(100-x)=y(y-1)$
The right side is even. Modulo $5$ the left side is $\in\{0,1,2\}$ and the right side is $\in\{0,1,4\}$. Thus both sides end with $0$ or $6$. In addition the left side must be $\in\{0,1\}\bmod3$ and the right side $\in\{0,2\}\bmod3$, so both sides are multiples of $3$.
So we need $x$ to be a multiple of $3$ or one more than such, $x(100-x)$ to end in $0$ or $6$ (forcing $x$ to end in one of the digits $0,2,8$), and $4[x(100-x)]+1=(2y-1)^2$ to be a perfect square. We set out to try eligible two-digit values of $x$ from $10$ up to $50$ (nine trials), knowing that any solution for $x$ in this range will be accompanied by a solution $100-x$ by symmetry of the product $x(100-x)$ on the left side. Blocks are cases where we do not get the needed perfect square for $4[x(100-x)]+1=(2y-1)^2$:
$x=10, x(100-x)=900, 4×900+1=3601■$
$x=12, x(100-x)=1056, 4×1056+1=4225=65^2☆$
$x=18, x(100-x)=1476, 4×1476+1=5905■$
$x=22, x(100-x)=1716, 4×1716+1=6865■$
$x=28, x(100-x)=2016, 4×2016+1=8065■$
$x=30, x(100-x)=2100, 4×2100+1=8401■$
$x=40, x(100-x)=2400, 4×2400+1=9601■$
$x=42, x(100-x)=2436, 4×2436+1=9745■$
$x=48, x(100-x)=2496, 4×2496+1=9985■$
So only $x=12$ works among numbers below $50$, from which $100-x=88$ will also work among numbers above $50$, and for this pair of solutions we see $2y-1=65, y=33$. Hence
$1233=12^2+33^2$
$8833=88^2+33^2$
Best Answer
We can minimize trial and error with some clever use of modular arithmetic.
Let $N=100a+10c+e$ be the square root. Thus $N^2\equiv e^2$ and we require also $N^2\equiv e\bmod 10$. Therefore $e^2\equiv e$ forcing $e\in\{0,1,5,6\}$.
We also know that $(100a)^2<10000(a+1)$ or $a^2<a+1$ forcing $a=1$. Then $N^2<20000$ but $145^2>140×150=21000$, therefore $N<145$. This result together with the earlier constraint on $e$ leaves only eighteen candidates, which can be exhaustively searched with little trouble; but we can do even better than that.
Consider the case $e=0$. Then $N=100+10c$ (with $a=1$) and $N^2=10000+2000c+100c^2$. For the hundreds digit in $N^2$ to match $c$ we must then have $c^2\equiv c\bmod 10$. This constraint admits$c\in\{0,1,5,6\}$, but only $0$ and $1$ satisfy the bounty $N<145$ which implies $c\le 4$. Thereby we identify
$100^2=10000$
$110^2=12100$
For $e=1$ we have
$N^2=10000+2000c+100(c^2+2)+20c+1$
With $c\le 4$, $20c+1<100$ and thus the hundreds digit is $\equiv c^2+2\bmod 10$. Therefore we must satisfy
$c^2-c+2\equiv 0\bmod 10$
which has a discriminant that is not a quadratic residue $\bmod 5$. So nobody's home here.
The cases $e=5$ and $e=6$ are left to the reader; they are handled similarly to $e=1$ as described above. For these cases $N<145$ implies $c\le 3$ which will then fix the hundreds digit of $N^2$ as $\equiv c^2+c$ ($e=5$) or $\equiv c^2+c+2$ ($e=6$). We will then get only one additional solution which the reader can find. I list the complete solution set as (with $x$ digits to be filled in):
$100^2=10000$
$1xx^2=1xxxx$
$110^2=12100$