I have seen method 2 used more often in stackoverflow, but I prefer method 1.
Suggestion: prefer method 2.
Both methods work with single functions. The problem arises when you have more than a function, with the same signature, and you want enable only one function of the set.
Suppose that you want enable foo(), version 1, when bar<T>() (pretend it’s a constexpr function) is true, and foo(), version 2, when bar<T>() is false.
With
template <typename T, typename = std::enable_if_t<true == bar<T>()>>
void foo () // version 1
{ }
template <typename T, typename = std::enable_if_t<false == bar<T>()>>
void foo () // version 2
{ }
you get a compilation error because you have an ambiguity: two foo() functions with the same signature (a default template parameter doesn’t change the signature).
But the following solution
template <typename T, std::enable_if_t<true == bar<T>(), bool> = true>
void foo () // version 1
{ }
template <typename T, std::enable_if_t<false == bar<T>(), bool> = true>
void foo () // version 2
{ }
works, because SFINAE modify the signature of the functions.
Unrelated observation: there is also a third method: enable/disable the return type (except for class/struct constructors, obviously)
template <typename T>
std::enable_if_t<true == bar<T>()> foo () // version 1
{ }
template <typename T>
std::enable_if_t<false == bar<T>()> foo () // version 2
{ }
As method 2, method 3 is compatible with selection of alternative functions with same signature.