Partial template specialization on noexcept? [duplicate]

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• How can I detect whether a template argument is a noexcept function? 2 answers

I'm working on a variant of std::function. One thing I'd like to be able to do is to allow people to specify the function signature as Function<void() noexcept> and have that only accept functors and function pointers that are noexcept callables.

I tried to do the following, but it didn't compile:

template<typename T>
struct Function;

template<typename R, typename... Args>
struct Function<R(Args...)> {};

template<typename R, typename... Args>
struct Function<R(Args...) noexcept> {};

How can I achieve the desired result? That the instantiation Function<void() noexcept> is distinct from Function<void()>.

One easy alternative I already know about is to just add on another template parameter, say template<typename T, bool IS_NOEXCEPT = false>. I'd rather not do that. I'd rather have the syntax of Function<void() noexcept>, not Function<void(), true>.

Note: even if we are talking C++17, I'm pretty sure the is_nothrow_callable type trait won't cut it.