You could order by numLogins
and take the first result, but that won't work if multiple people have the same number. Instead, find the max number, then find all users with that number.
from sqlalchemy import func
max_logins = db.session.query(func.max(User.numLogins)).scalar()
users = db.session.query(User).filter(User.numLogins == max_logins).all()
You could reduce this to one query.
sub = db.session.query(func.max(User.numLogins).label('ml')).subquery()
users = db.session.query(User).join(sub, sub.c.ml == User.numLogins).all()