Using back references, this problem becomes fairly easy (and thankfully js regex supports this).

The regex below matches only the invalid inputs.

^(\d)\1+-\1$

Regex Demo

^     - Start of line
(\d)  - create a capturing group in the first digit (important for the next part to work)
\1+   - back reference group 1 (\d+) and match it one or more times
-     - match a hyphen
\1    - back reference group 1
$     - end of line 

Then for the JS side, if the regex matches you know the input was invalid.

const inputs = [
'0000-0',
'0000-1',
'1111-1',
'1234-2'
];

const regex = /^(\d)\1+-\1$/;

inputs.forEach(item => {
 if (item.match(regex)) {
    console.log(item + ' - INPUT NOT ALLOWED')
 } else {
    console.log(item + ' - VALID INPUT')
 } 
});

JS Demo

EDIT

To handle the new requirements, you can just or the regex to another regex checking the entire length.

^(?:(\d)\1+-\1$|.{7,})

Demo

And since it sounds like you're using Java:

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Main {

    public static void main(String[] args) {
        // String to be scanned to find the pattern.
        String[] lines = {
                "0000-0",
                "0000-1",
                "1111-1",
                "1234-2",
                "1234567-2",
                "1234-26"
        };

        String pattern = "^(?:(\\d)\\1+-\\1$|.{7,})";
        // Create a Pattern object
        Pattern r = Pattern.compile(pattern);

        for(String line : lines) {
            System.out.print(line);
            Matcher m = r.matcher(line);
            if (m.find()) {
                System.out.println(" - INVALID");
            }else {
                System.out.println(" - VALID");
            }
        }
    }
}

** Output **

0000-0 - INVALID
0000-1 - VALID
1111-1 - INVALID
1234-2 - VALID
1234567-2 - INVALID
1234-26 - INVALID

Process finished with exit code 0