Ignore text in double brackets with batch replacement

I have a list of words (in a linked file) which I want to run to a text and have replaced any instances to wikilinks, i.e. find "text" and replace with "[[text]]". However, the existing text already contains words and phrases with wikilinks, they could be like these:

[[test]]
look at my [[test|testing]] how it tests
I have a test, you know a [[test, and more test]]
[[this is another test you know]]

Now, if the word is enclosed in brackets as listed above, then it should not be replaced. So, in the above example, the output should be for the "text" replacement:

[[test]]
look at my [[test|testing]] how it tests
I have a [[test]], you know a [[test, and more test]]
[[there's another test you know]]

So, only one replacement in line 3. The only way I thought of doing this is first to add an extra non-occurring character to words contained within double brackets. I.e first do something like

[[test한]]
look at my [[test한|testing한]] how it tests
I have a test, you know a [[test한, and한 more한 test한]]
[[there한's한 another한 test한 you한 know한]]

And then run the replacements file. But I cannot figure out the regex to add the non-occurring character as described. Any thoughts on this approach, or any other indeed, would be appreciated.

You can use:

(?<!\[)test(?!\])

(?!pattern) Negative Lookahead (?<!pattern) Negative Lookbehind

Thanks! Just tried with the test phrase but it also matches words within a phrase contained in double brackets, ie

[[a word, test, and more test]]

To keep it simple (and leave the find/replace linked file as is, ie. test→[[test]]) is it possible to add the extra character in any words between double brackets as mentioned, ie find: I have a [[test]], you know a [[test, and more test]] Replace I have a test, you know a [[test한, and한 more한 test한]]

Correct replace should be: I have a [[test한]], you know a [[test한, and한 more한 test한]]

I think I got it: Find: (?:\G(?!\A)|\[\[)(?:(?!\[\[|]]).)*?\K\w+ Replace: $0한