Proving that expression can be expressed as a sum of $3n-1$ natural square numbers that are bigger than one $n$ is a natural number. Prove that
$(2n+1)^3 - 2$ can be expressed as sum of $3n - 1$ natural square numbers, which are bigger than one.
Is this a valid proof?
Showing that $((4+1)^3 - 2) - ((2+1)^3 - 2)=98= 3^2 + 5^2 + 8^2$  can be expressed as a sum of $(6-1)-(3-1) = 3$ natural square numbers that are bigger than one. Then showing for $n=k+1$ and $n=k$, that it can be expressed as $(3(k+1) - 1) - (3k-1) = 3$ natural square numbers?

You have made a good start on using induction to prove the hypothesis that
$$p(n) = (2n + 1)^3 - 2 \tag{1}\label{eq1A}$$
can be expressed as the sum of $3n - 1$ perfect squares, each greater than $1$. First, the base case is $n = 1$, where $p(1) = 3^3 - 2 = 25$, which can be expressed as the sum of $3n - 1 = 2$ perfect squares as $25 = 9 + 16 = 3^2 + 4^2$.
Next, assume the hypothesis is true for $n = k$ for some $k \ge 1$. Then, as you indicated, showing that $p(k+1) - p(k)$ can be expressed as the sum of $3$ squares means these squares can be added to the $3k - 1$ squares already used for $p(k)$, thus showing that $p(k + 1)$ is a sum of $(3k - 1) + 3 = 3(k + 1) - 1$ perfect squares. Note that
$$\begin{equation}\begin{aligned}
p(k + 1) - p(k) & = ((2(k + 1) + 1)^3 - 2) - ((2k + 1)^3 - 2) \\
& = (2k + 3)^3 - (2k + 1)^3 \\
& = 8k^3 + 3(4)(3)k^2 + 3(2)(9)k + 27 - (8k^3 + 3(4)k^2 + 3(2)k + 1) \\
& = 8k^3 + 36k^2 + 54k + 27 - (8k^3 + 12k^2 + 6k + 1) \\
& = 24k^2 + 48k + 26 \\
& = (4k^2 + 4k + 1) + (4k^2 + 12k + 9) + (16k^2 + 32k + 16) \\
& = (2k + 1)^2 + (2k + 3)^2 + (4k + 4)^2
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Since $k \ge 1$, each perfect square above is $\gt 1$. As explained earlier, this means the hypothesis is also true for $n = k + 1$ so, by induction, it's true for all $n \ge 1$.
