Solving integer equation $a^3+b^3=a^2+72ab+b^2$

Suppose $d = (a, b)$, $a = da_0$, $b = db_0$, then $(a_0, b_0) = 1$ and$$ a^3 + b^3 = a^2 + 72ab + b^3 \Longleftrightarrow d(a_0 + b_0)(a_0^2 - a_0 b_0 + b_0^2) = a_0^2 + 72a_0 b_0 + b_0^2, \tag{1} $$ which implies $a_0 + b_0 \mid a_0^2 + 72a_0 b_0 + b_0^2$. Because$$ (a_0^2 + 72a_0 b_0 + b_0^2) - (a_0 + b_0)(a_0 + 71b_0) = -70b_0^2 $$ and $(a_0 + b_0, b_0) = 1$, then $a_0 + b_0 \mid 70$.

From (1) there is also $a_0^2 - a_0 b_0 + b_0^2 \mid a_0^2 + 72a_0 b_0 + b_0^2$. Because$$ (a_0^2 + 72a_0 b_0 + b_0^2) - (a_0^2 - a_0 b_0 + b_0^2) = 73a_0 b_0, $$ and $(a_0^2 - a_0 b_0 + b_0^2, a_0) = (b_0^2, a_0) = 1$, analogously $(a_0^2 - a_0 b_0 + b_0^2, b_0) = 1$, then$$ a_0^2 - a_0 b_0 + b_0^2 \mid 73 \Longrightarrow a_0^2 - a_0 b_0 + b_0^2 = 1 \text{ or } 73. $$

If $a_0^2 - a_0 b_0 + b_0^2 = 1$, then$$ 1 = a_0^2 - a_0 b_0 + b_0^2 \geqslant \frac{1}{2} (a_0^2 + b_0^2) \Longrightarrow a_0^2 + b_0^2 \leqslant 2 \Longrightarrow a_0 = b_0 = 1,\ d = 37, $$ thus $(a, b) = (37, 37)$. If $a_0^2 - a_0 b_0 + b_0^2 = 73$, denote $s = a_0 + b_0$, then$$ \frac{s^2}{4} \leqslant a_0^2 - a_0 b_0 + b_0^2 = 73 < s^2 \Longrightarrow 8 < s < 18. $$ Note that $s \mid 70$, thus $s = 10$ or $14$. If $s = 10$, then$$ a_0^2 - a_0(10 - a_0) + (10 - a_0)^2 = 73 \Longrightarrow a_0 = 1 \text{ or } 9,\ d = 1, $$ thus $(a, b) = (1, 9)$ or $(9, 1)$. If $s = 14$, then$$ a_0^2 - a_0(14 - a_0) + (14 - a_0)^2 = 73 \Longrightarrow a_0 = 7 \pm \sqrt{22}, $$ a contradiction. Therefore, $(a, b) = (1, 9),\ (9, 1),\ (37, 37)$.

Let $a=x+y, b=x-y$. Then: $$a^3+b^3=a^2+72ab+b^2 \Rightarrow \\ (x+y)^3+(x-y)^3=(x+y)^2+72(x+y)(x-y)+(x-y)^2 \Rightarrow \\ 2x^3+6xy^2=2x^2+72x^2 -72y^2+2y^2 \Rightarrow \\ y^2=\frac{(37-x)x^2}{35+3x}\ge 0.$$ Note that: $$x=\frac{a+b}{2}>0 \ \ \text{and} \ \ 37-x\ge0 \Rightarrow 0<x\le 37.$$ One can quickly check (if not further analysis) all numbers to find: $(x,y)=(5,4),(37,0)$.

Hence: $$(a,b)=(9,1),(37,37) \ \ \text{and} \ \ (1,9) \ \text{(due to symmetry)}.$$