Since this is number theory I believe you meant natural numbers N.

If $ N$ is divisible by 10, it is divisible by 100, so keep dividing through by 100 until it is not divisible by 10, so that we finally end up with $ N = 3000..0a$ or $ a000..003$, with $ a$ an integer, $ 1\le a\le 9$. If there are no 0's, then only $ 36$ works. Otherwise, assume there is at least 1 0 inbetween $ a$ and $ 3$. The only square residues mod 9 are $ 0,1,4,7$, so the only possible values of $ a$ are $ 6,7,1,4$ since $ N$ mod 9 is $ 3 + a$. mod 4, $ a000..003$ is congruent to 3, so is not a square. Therefore, the only possible candidates are $ 3000..01$, $ 3000..04$, $ 3000..06$, $ 3000..07$. However, the one ending in 6 and the one ending in 7 don't work mod 4, so only the first 2 are possible candidates. $ 3*10^{n} = x^2 - 1 = (x - 1)(x + 1)$, boring casework should rule this out, since $ gcd(x - 1,x + 1) = 1,2$, similarly for the other one, since $ gcd(x - 2,x + 2) = 1,2,4$, someone who is not as tired, or who enjoys doing such casework can complete the solution . Therefore $ 36*100^{n}$ are the only integers that satisfy it, for $ n$ being any integer $ \ge 0$.

Rofler wrote: $ 3*10^{n} = x^2 - 1 = (x - 1)(x + 1)$, boring casework should rule this out, since $ gcd(x - 1,x + 1) = 1,2$, similarly for the other one, since $ gcd(x - 2,x + 2) = 1,2,4$, someone who is not as tired, or who enjoys doing such casework can complete the solution . I solved this recently. Started the same way you did. I didn't use GCD but still completed the solution, don't know if this is correct, but here's what I did: $3 \cdot 10^n$ = (x-1)(x+1) or (x-2)(x+2), so I thought of splitting $2^n \cdot 3 \cdot 5^n$ into two sets of factors which are either 2 apart or 4 apart. Since they are 2 or 4 apart they both need to be even (or odd but that's not possible because the product would be odd) so I put a two into each of them. Now if we put a 5 into each of them then both numbers will be divisible by 10 and cannot be 2 or 4 apart, so that case does not work, therefore we need to put every single 5 into one of the factors, and that would leave it way larger than the other factor and therefore more than 2 or 4 apart (easy to prove). It feels too simple but it might work.