The reason it works for 6 and 66 is that 6=3*4/2, so the 3*4 is 12 and the triangular number 66 is that same 12 times 11 divided by 2.   Concatenating a one digit number is the same as multiplying by 11.

To concatenate an n-digit number you need a "concatenation generating multiplier", or CGN of the form 1z1, where 'z' represents a string of n-1 zeros.   So 123*1001 = 123123

We would need to find either a CGN+1 or CGN-1 which is double a triangular number.  This works for CGN=11.  12 is double a triangular number:  2*6.

Using the quadratic equation on n(n+1)/2 = T yields a determinant:  √(1+8T).

So a requirement for T to be triangular is that (CGN±1)*4 + 1 must be a perfect square.

The possible values for this expression are:

41, 49, 401, 409, 4001, 4009, ...

As these number get larger, half of them are slightly more than 4 followed by an even number of zeros, so being just 1 or 9 above a perfect square can not be a perfect square.

The other half are either 1 or 9 greater than 40 followed by an even number of zeros.  Since √40 is irrational (here is the hand waving argument which makes it not quite a proof) it is unlikely that there will be an integer which is a tiny amount greater than  √40 * 10^k