ID generator with no matching contiguous pairs

Brachylog, 11 10 bytes

{≤ṫẹ~ḅẉ}ᶠ⁾

Input is a list of two numbers. Try it online!

Explanation

The built-in ḅ takes a list or string, like "1000220", and splits it into blocks of equal adjacent elements, like ["1","000","22","0"]. In this program, I apply the ~ operator to it, so it works in reverse: it takes a list of strings, checks that each string consists of repetitions of a single character and neighboring strings have different characters, and concatenates the list. The predicate ≤ enumerates numbers starting from the first input in increasing order, and I check a condition on them, printing those that satisfy it and stopping when I've found enough.

{≤ṫẹ~ḅẉ}ᶠ⁾  Input is a pair, say [M=988,N=3].
{      }ᶠ⁾  Apply this predicate to M and compute the first N results.
 ≤          Take a number that is greater than or equal to M (first up is 988).
  ṫ         Convert it to string: "988"
   ẹ        Split it into substrings of length 1: ["9","8","8"]
     ~ḅ     Apply ḅ in reverse: fails, try next number.
       ẉ    If ḅ succeeds, print the resulting string and a newline.
            This counts as a result of the predicate.

05AB1E, 9 bytes

µÐÔQi=¼}>

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Exlpanation

µ           # loop until counter equals n
 Ð          # triplicate current value (initially m)
  Ô         # connected uniqueified on the copy at the top of the stack
   Q        # compare to unmodified for equality
    i  }    # if true
     =      # print current value while leaving it on the stack
      ¼     # increment the counter
        >   # increment current value

Java 8, 83 bytes

(m,n)->{for(;n>0;m++)if(!(m+"").matches(".*(.)\\1.*")){System.out.println(m);n--;}}

Explanation:

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(m,n)->{                      // Method with two integer parameters and no return-type
  for(;n>0;                   //  Loop as long as `n` is larger than 0
      m++)                    //    After every iteration: increase `m` by 1
    if(!(m+"").matches(".*(.)\\1.*")){
                              //   If there are no repeated adjacent digits:
      System.out.println(m);  //    Print `m`
      n--;}}                  //    And decrease `n` by 1