Charcoal, 40 bytes

Ｎθ≔¹ηＷ¬№υθ«≦⊕η≔∨υ⟦⁰⟧ζ≔⟦⟧υＦ⊕θＦζ⊞υ⁺λＸκη»Ｉη

Try it online! Link is to verbose version of code. Outputs the nth value. Brute force, so times out on TIO for n=23 and similar higher values. Explanation:

Ｎθ

Input n.

≔¹η

Start searching for powers.

Ｗ¬№υθ«

Repeat until a one of the sums includes the input.

≦⊕η

Try the next power.

≔∨υ⟦⁰⟧ζ

Save the previous set of sums, or use a sum of zero for the first pass (this is needed to make the code output 2 for an input of zero instead of the 1 that an empty sum would imply.)

≔⟦⟧υ

Start a new set of sums.

Ｆ⊕θ

Loop over all of the integers up to and including n.

Ｆζ

Loop over all of the previous sums.

⊞υ⁺λＸκη

Add the current power to the previous sum and save the result.

»Ｉη

Output the highest power needed.

I wasn't able to golf this much more efficient version below 41 bytes:

≔ＥＮ⊗¬ιθＦ…²χＦ⮌⌕Ａθ⁰Ｆ⊙θ∧¬‹κＸμι§θ⁻κＸμι§≔θκιＩθ

Try it online! Link is to verbose version of code. Outputs the first n values. Explanation:

≔ＥＮ⊗¬ιθ

Start with 0 having been found to have a value of 2 but none of the other values having been found yet.

Ｆ…²χ

Try from squares up to ninth powers.

Ｆ⮌⌕Ａθ⁰

Loop over all of the values that haven't yet been found in reverse order, so the highest index is tested first.

Ｆ⊙θ∧¬‹κＸμι§θ⁻κＸμι

Subtract the powers of all the integers up to n from the current index and see if any of those value had previously been found.

§≔θκι

If so then mark this value with the necessary power.

Ｉθ

Output all of the values.

Charcoal, 40 39 bytes

ＦＮ⊞υ⊗¬ιＦ…²χＦ⮌⌕Ａυ⁰Ｆ⊙υΣ✂…υ⊕⁻κＸμι±¹§≔υκιＩυ

Try it online! Link is to verbose version of code. Outputs the first n values. Explanation:

ＦＮ⊞υ⊗¬ι

Start with 0 having been found to have a value of 2 but none of the other values having been found yet.

Ｆ…²χ

Try from squares up to ninth powers.

Ｆ⮌⌕Ａυ⁰

Loop over all of the values that haven't yet been found in reverse order, so the highest index is tested first.

Ｆ⊙υΣ✂…υ⊕⁻κＸμι±¹

Subtract the powers of all the integers up to n from the current index and see if any of those values had previously been found. This is harder than it sounds as it's necessary to defeat Charcoal's cyclic indexing.

§≔υκι

If so then mark this value with the necessary power.

Ｉυ

Output all of the values.

Charcoal, 40 bytes

Ｎθ≔¹ηＷ¬№υθ«≦⊕η≔∨υ⟦⁰⟧ζ≔⟦⟧υＦ⊕θＦζ⊞υ⁺λＸκη»Ｉη

Try it online! Link is to verbose version of code. Outputs the nth value. Brute force, so times out on TIO for n=23 and similar higher values. Explanation:

Ｎθ

Input n.

≔¹η

Start searching for powers.

Ｗ¬№υθ«

Repeat until a one of the sums includes the input.

≦⊕η

Try the next power.

≔∨υ⟦⁰⟧ζ

Save the previous set of sums, or use a sum of zero for the first pass (this is needed to make the code output 2 for an input of zero instead of the 1 that an empty sum would imply.)

≔⟦⟧υ

Start a new set of sums.

Ｆ⊕θ

Loop over all of the integers up to and including n.

Ｆζ

Loop over all of the previous sums.

⊞υ⁺λＸκη

Add the current power to the previous sum and save the result.

»Ｉη

Output the highest power needed.

Charcoal, 40 bytes

Ｎθ≔¹ηＷ¬№υθ«≦⊕η≔∨υ⟦⁰⟧ζ≔⟦⟧υＦ⊕θＦζ⊞υ⁺λＸκη»Ｉη

Try it online! Link is to verbose version of code. Outputs the nth value. Brute force, so times out on TIO for n=23 and similar higher values. Explanation:

Ｎθ

Input n.

≔¹η

Start searching for powers.

Ｗ¬№υθ«

Repeat until a one of the sums includes the input.

≦⊕η

Try the next power.

≔∨υ⟦⁰⟧ζ

Save the previous set of sums, or use a sum of zero for the first pass (this is needed to make the code output 2 for an input of zero instead of the 1 that an empty sum would imply.)

≔⟦⟧υ

Start a new set of sums.

Ｆ⊕θ

Loop over all of the integers up to and including n.

Ｆζ

Loop over all of the previous sums.

⊞υ⁺λＸκη

Add the current power to the previous sum and save the result.

»Ｉη

Output the highest power needed.

I wasn't able to golf this much more efficient version below 41 bytes:

≔ＥＮ⊗¬ιθＦ…²χＦ⮌⌕Ａθ⁰Ｆ⊙θ∧¬‹κＸμι§θ⁻κＸμι§≔θκιＩθ

Try it online! Link is to verbose version of code. Outputs the first n values. Explanation:

≔ＥＮ⊗¬ιθ

Start with 0 having been found to have a value of 2 but none of the other values having been found yet.

Ｆ…²χ

Try from squares up to ninth powers.

Ｆ⮌⌕Ａθ⁰

Loop over all of the values that haven't yet been found in reverse order, so the highest index is tested first.

Ｆ⊙θ∧¬‹κＸμι§θ⁻κＸμι

Subtract the powers of all the integers up to n from the current index and see if any of those value had previously been found.

§≔θκι

If so then mark this value with the necessary power.

Ｉθ

Output all of the values.