Django IntegerChoices with variable label

I have a class that inherits from models.IntegerChoices and defines certain constants I use in my model. I want to make the "labels" of my integer choices a variable in the settings.py:

class Status(models.IntegerChoices):
    ONE = 0, settings.ONE_LABEL
    TWO = 1, settings.TWO_LABEL

When playing around with this I found that whenever I change ONE_LABEL or TWO_LABEL in my settings, Django would like to create a new migration similar to

migrations.AlterField(
    model_name='mymodel',
    name='status',
    field=models.IntegerField(choices=[(0, 'New one label'), (1, 'New two label')], default=0),
)

My goal is to deploy multiple (very similar) servers where only UI elements (that display the names of status) change according to the values in my settings.
I guess it is not wise to allow changes in my settings to trigger any changes in the database, as that would make updating later versions of my project much harder. Is it safe to ignore the possible migration?
I thought the way above is the easiest as I have multiple model forms and other components that use the label of my status class. This would allow a single change in my settings to be transferred throughout my project. I can't see why changing the label would require a DB migration as long as I don't change the values. I wonder if there is any better way to achieve the same goal.

Django create new migrations when it detects changes between your current models in your code and your migrations.

To avoid django from creating a new migration when you change your settings, you could update your migration file directly, so that no changes would be detected :

from django.db import migrations, models
from django.conf import settings

class Migration(migrations.Migration):

    dependencies = [
        # your dependencies
    ]

    operations = [
        migrations.AlterField(
            model_name='mymodel',
            name='status',
            field=models.IntegerField(choices=[(0, settings.ONE_LABEL), (1, settings.TWO_LABEL)], default=0),
)
    ]