While investigating $a^n - b^n$, I obtained conditions on $n$ for $a^2 - b^2 = n$.
However, starting from the 3rd power, the problem becomes significantly more complex.
Although something can still be said based on:
$$ \sum_\limits{k=1}^x \left( k^n - (k-1)^n \right)=x^n $$
To represent a number as a difference of cubes, it must be a consecutive sum of numbers of the form $6 \left(\frac{x(x-1)}{2}\right)+1$
$$6 \left(1\right)+1 + 6 \left(1+2\right)+ 1 =7+19 = 26 \Rightarrow3^3 - 1^3 = 26$$
Among the solutions, even squares can be found $8^3 - 7^3 = 13^2$.
I have found almost nothing about this, perhaps you will have good recommendations and ideas.
By $FLT$ for exponent $3$, the integer $n$ cannot be a cube so $n=m^3A$ where $A$ is a cube-free integer and your problem becomes about the integer points of the elliptic curve of equation $$x^3+y^3=Az^3\hspace0.9cm (*)$$ where $n=Az^3$ is given and you search about $x=a$ and $y=-b$. This is a difficult problem on which the Norwegian mathematician E. S. Selmer was the first to try deeply. Currently many values of $A$ are known that can be represented by $(*)$ but for the vast majority of these values of $A$ it is not known whether they are representable or not.
I have easily proved longtime ago some characterizations of $A$ in $(*)$ to be solvable in integers $x,y,z$. We have:
$(1)$ If $a+b=c$ then $A=abc$ is representable.
$(2)$ All product $A$ of two consecutive numbers is representable.
$(3)$ If $a+b=h^3$ then $A=ab$ is representable.
All this is a consequence of the identity $$a+b=c\iff [9ab^2+(b-a)^3]^3+[9a^2b-(b-a)^3]^3=abc[3(a^2+ab+b^2)]^3$$
You can check as many examples as you want with a calculator.