Make Iterators.peel return nothing when itr empty

Fixes JuliaLang#39569.

Old behaviour:
```jl
try
    h, t = Iterators.peel(itr)
    # do something with h and t
catch e
    if e isa BoundsError
        # itr was probably empty
    else
        rethrow()
    end
end
```

New behaviour:
```jl
x = Iterators.peel(itr)
if isnothing(x)
    # itr was empty
end
h, t = x
```

NEWS.md

@@ -10,6 +10,7 @@ New language features
 Language changes
 ----------------
 
+* `Iterators.peel(itr)` now returns `nothing` when `itr` is empty rather than throwing a `BoundsError`. ([#39569])
 
 Compiler/Runtime improvements
 -----------------------------

base/iterators.jl

@@ -556,6 +556,8 @@ rest(itr) = itr
 
 Returns the first element and an iterator over the remaining elements.
 
+If the iterator is empty, returns `nothing` (like `iterate`).
+
 # Examples
 ```jldoctest
 julia> (a, rest) = Iterators.peel("abc");
@@ -571,7 +573,7 @@ julia> collect(rest)
 """
 function peel(itr)
     y = iterate(itr)
-    y === nothing && throw(BoundsError())
+    y === nothing && return y
     val, s = y
     val, rest(itr, s)
 end

test/iterators.jl

@@ -848,3 +848,11 @@ end
     @test cumprod(x + 1 for x in 1:3) == [2, 6, 24]
     @test accumulate(+, (x^2 for x in 1:3); init=100) == [101, 105, 114]
 end
+
+@testset "Iterators.peel" begin
+    @test Iterators.peel([]) == nothing
+    @test Iterators.peel(1:10)[1] == 1
+    @test Iterators.peel(1:10)[2] |> collect == 2:10
+    @test Iterators.peel(x^2 for x in 2:4)[1] == 4
+    @test Iterators.peel(x^2 for x in 2:4)[2] |> collect == [9, 16]
+end