Determine if an Array contains something other than 2

Python 2, 43 40 bytes

f=lambda l:l>=[]and all(map(f,l))or l==2

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At time of posting this answer, it was still allowed per this meta consensus to output via throwing an error / not throwing an error. Therefore this answer at 26 bytes was valid:

f=lambda l:l==2or map(f,l)

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Prolog (SWI), 43 33 bytes

I smell... recursion.

Thanks to Emigna and Leaky Nun for saving 10 bytes!

Code

a([]).
a([X|T]):-(X=2;a(X)),a(T).

Try it online! or Verify all test cases!

Explanation:

For non-Prolog users, a list is formatted in the following way: [Head | Tail].

The Head is the first element of the list, and tail is the remaining list. Test it here!. An important case here is that the tail of a list with 1 element is equal to []. You can test that here.

% State that an empty array is truthy.
a([]).

% If the list is not empty (covered by the previous line), we need to check
% whether the Head is equal to 2 or whether the head is truthy.
% After that, we only need to check if the remaining list is truthy.
a([Head | Tail]) :- (Head = 2; a(Head)), a(Tail).

Jelly, 4 bytes

F;2E

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How it works

F;2E
F    flatten
 ;2  append 2
   E all elements are equal