First, you show that you can express all numbers $129$ though something convenient as a sum of distinct squares just by making a list. I will choose $4900$ and assume that we have a list that runs that high. We do the proof by strong induction. Assume that all numbers in the range $[129,k]$ can be expressed as a sum of distinct squares, where $k \ge 4900$. We shall show that $k+1$ can be so expressed. Let $m = \lfloor \sqrt {k+1} \rfloor -1$. We can express $k+1=m^2+p$, where $p \ge 129$, so it can be expressed as a sum of distinct squares. Since $p \lt m^2$, there is no conflict with the new square, and $k+1$ can be expressed as a sum of distinct squares. You could lower the bound of the specific list below $3600$ if you think harder about what square to subtract. I chose $3600$ because the spacing of squares attains $129$ at that point.
Added: the list you make can run only through $[129,297]$. We can be lazier in the first part with more work in the second. $297$ is a risk because $297=169+128$, so the largest square you can subtract and stay larger than $129$ is $144$ and the expression for $153$ could involve $144$. In fact $144+9=153$, but you can also do $100+49+4$. Starting with $298$ you can always subtract a square that leaves at least $129$ and is greater than half the starting number, so the strong induction goes through.
In the paper Characterizing the Sum of Two Cubes, Kevin Broughan gives the relevant theorem,
Theorem: Let $N$ be a positive integer. Then the equation $N = x^3 + y^3$ has a solution in positive integers $x,y$ if and only if the following conditions are satisfied:
There exists a divisor $m|N$ with $N^{1/3}\leq m \leq (4N)^{1/3}.$
And $\sqrt{m^2-4\frac{m^2-N/m}{3}}$ is an integer.
The sequence of integers $F(n)$,
$$\begin{aligned}
F(n)
&= a^3+b^3 = (2 n + 6 n^2 + 6 n^3 + n^4)^3 + (n + 3 n^2 + 3 n^3 + 2 n^4)^3\\
&= c^3+d^3 = (1 + 4 n + 6 n^2 + 5 n^3 + 2 n^4)^3 + (-1 - 4 n - 6 n^2 - 2 n^3 + n^4)^3
\end{aligned}$$
for integer $n>3$ apparently is expressible as a sum of two positive integer cubes in exactly and only two ways.
$$\begin{aligned}
F(4) &= 744^3+756^3 = 945^3+15^3\\
F(5) &= 1535^3+1705^3 = 2046^3+204^3\\
&\;\vdots\\
F(60) &=14277720^3+26578860^3 = 27021841^3+12506159^3
\end{aligned}$$
Using Broughan's theorem, I have tested $F(n)$ from $n=4-60$ and, per $n$, it has only two solutions $m$, implying in that range it is a sum of two cubes in only two ways. Can somebody with a faster computer and better code test it for a higher range and see when (if ever) the proposed statement breaks down? Incidentally, we have the nice relations,
$$a+b = 3n(n+1)^3$$
$$c+d = 3n^3(n+1)$$
Note: $F(60)$ is already much beyond the range of taxicab $T_3$ which is the smallest number that is the sum of two positive integer cubes in three ways.
$$T_3 \approx 444.01^3 = 167^3+436^3 = 228^3+423^3 = 255^3+414^3$$
(Using the theorem, this yields 3 values for $m$.)
Best Answer
Note that this problem is equivalent to the problem of partitioning $n$ into $4$ parts.
In number theory, a partition is one in which the order of the parts is unimportant. Under these conditions, we are under the illusion that the order is important, however we can only order any given combination of $4$ parts in $1$ way. Thus, the order is unimportant.
There is a well known recurrence for $p(n,k)$, the number of partitions of $n$ into $k$ parts. It is given by $$p(n,k) = k\cdotp(n-1,k) + p(n-1,k-1).$$ Your question is only concerned with the case of $k=4$.