Sum of two cubes equal to prime square If $a,b\in \mathbb{N}$ find all primes $p$ such that $a^3+b^3=p^2$
My approach-
$a^3+b^3=(a+b)(a^2-ab+b^2)=p^2$ suppose $a+b=x$ and $a^2-ab+b^2=y$ then there are two cases- $(x,y)=(p^2,1),(p,p)$
Now I am struggling for case 02 where $(x,y)=(p,p)$ 

I think there's a big mistake in the two cases conclusion.
We have $(x,y) \in \{(p^2,1)(p,p)(1,p^2)\}$
So let's divide them, (noting that $a,b \ne 0$)
Case 1:
$$a+b=a^2-ab+b^2=p$$
$$(a+b)^2-3ab=p \implies p^2-3ab=p \iff 3ab=p(p-1)$$
$$p=3 \implies ab=2 \implies (a,b) \in \{(1,2),(2,1)\}$$
$$p \ne3, 3| \ p(p-1) \implies p \equiv 1 \pmod 3 \implies p=3k+1 \implies ab=kp \tag{for some $k \in \mathbb{N^*}$}$$
$$a+b=p \iff a=p-b \implies b(p-b)=kp \iff b^2=p(b-k)$$
$$p\ |\ b^2 \implies p \ | \ b \implies b=mp \implies b^2=m^2p^2 \implies m^2p=mp-k $$
for some positive integer $m$, now this rewrites to
$$mp(1-m)=k$$
which is of course impossible for positive  $k$.
Case 2: $a+b=1$, $$a^2-ab+b^2=p^2 \iff (a+b)^2-3ab=p^2 \implies 1-3ab=p^2$$
which is, also, impossible
Case 3: $a+b=p^2$
$$a^2-ab+b^2=1 \iff (a+b)^2-3ab=1 \implies 3ab=p^4-1$$
$$a+b=p^2 \implies a=p^2-b \implies 3b(p^2-b)=p^4-1$$
$$\iff 3b^2-3bp^2+p^4-1=0$$
$$\implies b=\frac{1}{6} \left(3p^2 \pm \sqrt{9p^4-12(p^4-1)}\right)$$
and since the discriminant is less than zero, it's also impossible.
