Here is an indirect solution - a solution that doesn't use any specific primitive root. I think a better solution exists, but I would still like to post it because it gives a different perspective.
The solutions $1$ and $m-1$ are obviously always solutions. So the difficulty is proving there aren't any more.
By the primitive root theorem, the possibilities for $m$ are $m=2$, $m=4$, $m=p^k$, or $m=2p^k$ where $p$ is an odd prime and $k\ge 1$.
For $m=2$ we actually have $1=m-1$, and there is a unique square root of $1$.
For $m=4$ you may check directly that the claim holds, just by calculating.
For $m=p^k$, we prove by induction on $k$. For $k=1$ there are no zero divisors, so $(x-1)(x+1)=0$ implies $x-1=0$ or $x+1=0$. For $k>1$, assuming there are exactly two solutions modulo $p^{k-1}$, we use Hensel's lemma to claim that these lift to exactly two solutions modulo $p^k$.
For $m=2p^k$ we use the Chinese remainder theorem to combine the two solutions modulo $p^k$ with the unique solution modulo $2$ to get two solutions modulo $m$.
You can't reason from $256\mid ab$ to $256\mid a \lor 256\mid b$ because $256$ is not a prime. Instead you need to consider all the ways that the prime factors of $256$ can be distributed among $a$ and $b$.
Since $256$ is a prime power, namely $2^8$, there are only nine ways to write it as a product, and we can check those systematically:
Case 0. $256\mid(x-3)$ and $1\mid(x+3)$. This leads to the solution $x=3$.
Case 1. $128\mid(x-3)$ and $2\mid(x+3)$. The former implies the latter, so this leads to the solution $x=131$ as well as $x=3$ which we already know.
Case 2. $64\mid(x-3)$ and $4\mid(x+3)$. This would require the difference between $x-3$ and $x+3$ to be a multiple of $4$, which is never the case, so this is impossible.
Case 3. $32\mid(x-3)$ and $8\mid(x+3)$. Similar to case 2 -- impossible because $6$ is not a multiple of $8$.
Cases 4 through 6. Similar.
Case 7. $2\mid(x-3)$ and $128\mid(x+3)$. This gives $x=125$ or $x=253$.
Case 8. $1\mid(x-3)$ and $256\mid(x+3)$. This gives $x=253$, which we already know from case 7.
So the solutions are $\{3,125,131,253\}$.
Best Answer
As $(121,1800)=1,$ $$x^2\equiv121\pmod{1800}\iff(x\cdot 11^{-1})^2\equiv1$$
Now as $1800=3^2\cdot2^3\cdot5^2$
as $y^2\equiv1\pmod{p^n}$ has exactly two in-congruent solutions for odd prime $p$
and $y^2\equiv1\pmod8$ has four
the number of solutions $$(x\cdot 11^{-1})^2\equiv1\pmod{1800}$$ will be $2\cdot4\cdot2$