regex return false?

You were almost there. You must escape \d to \d within the RegExp constructor. Otherwise it is "forgotten". Notice here how the bar disappears:

console.log(new RegExp('^(max).(\d{1,3})'));

If you did regex so var pattMax = /'^(max).(\d{1,3})/; your code would almost work:

var pattMax = /^(max).(\d{1,3})/;
['max18', 'max120', 'max6'].forEach(function(rule) {
    console.log(pattMax.test(rule) ? 'é max seguido de 1 á 3 digitos' : 'Falhou!');
});

What fails is that it expects max plus any character except new line (the . ) and then numbers in the amount of 1 to 3, now it fails.

I think you could just use this: /^max\d{1,3}/ without constructor:

var pattMax = /^max\d{1,3}/;
['max18', 'max120', 'max6'].forEach(function(rule) {
    console.log(pattMax.test(rule) ? 'é max seguido de 1 á 3 digitos' : 'Falhou!');
});

You need to escape the \d or pass the argument i to the regex constructor.

var pattMax = new RegExp('^(max)(\d{1,3})', 'i');

pattMax.test(rule); //false?

if(pattMax.test(rule)){
    console.log('é max seguido de 1 á 3 digitos');
}

Based on: javascript new regexp from string

In theory, ^max\d{1,3}

The point in your original RegEx takes other characters, such as max_121.

Max120 would work for the wrong reason, the point would get 1, and /d{1,3} would get 20.

Describing your query :

^(max).(\d{1,3})
^                 marca decomeço da linha
 (   ) (       )  grupos de retorno
  max             literal "max"
      .           qualquer caractere, mas tem que ter algum
        \d{1,3}   de um a três dígitos

Proposed version

^max\d{1,3}
^                 começo da linha
 max              literal "max"
    \d{1,3}       de um a três digitos

Depending on usage, you can omit ^ (start of line), or even add $ at the end (end of line) if you are going to operate on larger strings and do not want to accept substrings.