The answer is no for all these questions. Take the line with infinite origins: the scheme obtained by gluing an infinite amount of copies of $\mathbb{A}^1$ along the open subsets $\mathbb{G}_m$. This has Krull dimension 1 (there are only closed points and the unique generic point) and it is irreducible (the only proper nonempty closed subsets are the closed points) but it is not quasi-compact (it contains an infinite discrete set), and so in particular not Noetherian.

If you want an example that's not even locally Noetherian you replace $\mathbb{A}^1$ with the nilpotent thickening obtained by adding an infinite amount of square-zero elements.