Revised and simplified version: Let $P(n)=n^5+n^3+2n^2+2n+2=p^3, Q(n)=2n^2+n+2=q^3$ $\Longrightarrow P(n)-Q(n)= n^5+n^3+n=p^3-q^3$ Taking mod $9\Longrightarrow n^5+n^3+n$ mod $9 \in$ {$0, 3,6$} (see below) $p^3,q^3$ mod $9 \in$ {$0, 1,8$}$\Longrightarrow p^3-q^3$ mod $9 \in$ {$0, 1,2, 7, 8$} Therefore, the only acceptable option is $n=9k$ for the LHS and $p=9l,q=9r$ or $p=9l+i,q=9r+i$ for the RHS However, replacing $n=9k$ in $n^5+n^3+n=p^3-q^3\Longrightarrow 9^5k^5+9^3k^3+9k=9^3l^3-9^3r^3$ (or $= (9l+i)^3-(9r+i)^3$) shows that the RHS is divisible by $9^3$ or $3^3$, unlike the LHS. Thus also this option fails and no $n$ satisfies the given requirements. $n=9k+1\longrightarrow n^5=1, n^3=1, n=1, n^5+n^3+n\equiv 3\bmod{9}$ $n=9k+2\longrightarrow n^5=5, n^3=8, n=2, n^5+n^3+n\equiv 6\bmod{9}$ $n=9k+3\longrightarrow n^5=0, n^3=0, n=3, n^5+n^3+n\equiv 3\bmod{9}$ $n=9k+4\longrightarrow n^5=7, n^3=1, n=4, n^5+n^3+n\equiv 3\bmod{9}$ $n=9k+5\longrightarrow n^5=2, n^3=8, n=5, n^5+n^3+n\equiv 6\bmod{9}$ $n=9k+6\longrightarrow n^5=0, n^3=0, n=6, n^5+n^3+n\equiv 6\bmod{9}$ $n=9k+7\longrightarrow n^5=4, n^3=1, n=7, n^5+n^3+n\equiv 3\bmod{9}$ $n=9k+8\longrightarrow n^5=8, n^3=8, n=8, n^5+n^3+n\equiv 6\bmod{9}$ .

Assume that $n^5 + n^3 + 2n^2 + 2n + 2 = a^3$ and $2n^2 + n + 2 = b^3$ for positive integers $a$ and $b$. Then $(a - b)(a^2 + ab + b^2) = a^3 - b^3 = n^5 + n^3 + n = n(n^2 - n + 1)(n^2 + n + 1)$. Note that RHS is divisible by $3$, so at least one of $a - b$ or $a^2 + ab + b^2 = (a - b)^2 + 3ab$ has to be divisible by $3$. Hence, both $a - b$ and $a^2 + ab + b^2$ are divisible by $3$, and $9 | n(n^2 - n + 1)(n^2 + n + 1)$. Since $n$, $n^2 - n + 1$, and $n^2 + n + 1$ are pairwise relatively prime, we have three cases: Case 1: $9 | n$. Then $2n^2 + n + 2 \equiv 2 \pmod{9}$, which cannot be a perfect cube. Case 2: $9 | n^2 - n + 1$. Then $n^2 - n \equiv 8 \pmod{9} \iff (2n-1)^2 \equiv 6 \pmod{9}$, which is impossible, since all perfect squares divisible by $3$ are divisible by $9$. Case 3: $9 | n^2 + n + 1$. Then $n^2 + n \equiv 8 \pmod{9} \iff (2n+1)^2 \equiv 6 \pmod{9}$, which is impossible, since all perfect squares divisible by $3$ are divisible by $9$. Thus, $n^5 + n^3 + 2n^2 + 2n + 2$ and $2n^2 + n + 2$ cannot be perfect cubes simultaneously.

Note that $n$ is clearly odd, as otherwise $n^5+n^3+2n^2+2n+2\equiv 2\pmod{4}$. Now, factorize this quantity as $(n+1)\left(n^4-n^3+2n^2+2\right)$; denote by $d$ the gcd of the factors. Then, $n\equiv -1\pmod{d}$ implies $n^4-n^3+2n^2+2\equiv 6\pmod{d}$, thus $d\in\{1,2,3,6\}$. As $n$ is odd, $d\in\{2,6\}$. Now, if $3\mid n+1$ then $n=3k-1$ for some $k$ for which $2n^2+n+2=18k^2-9k+3\equiv 3\pmod{9}$, thus it cannot be a perfect cube. Hence, $d=2$; leaving us with two cases. Case 1. $n+1=2a^3$. For $a=1$, $2n^2+n+2=5$ is not a cube. Let $a>1$. Using $n=2a^3-1$ we get $2n^2+n+2 = 8a^6-6a^3+3$, but $$\left(4a^3-1\right)^2<8a^6-6a^3+3<8a^6,$$ yielding a contradiction. Case 2. $n+1=4a^3$. In this case, $2n^2+n+2=32a^6-12a^3+3$. Note that $3\nmid a$, thus $a^6\equiv 1\pmod{9}$ and $a^3\equiv \pm 1\pmod{9}$ (by Euler). Hence, $2n^2+n+2\equiv 35\pm 12 \in\{2,5\}\pmod{9}$, none of which is a perfect cube.

Assume $2n^2 + n + 2$ is a cube. The fact that $2n^2 + n + 2 \equiv 2,5,3,5,2,3,3,8,3 \pmod{9}$ for $n = 0,1,2, \cdots, 8$, yields $n \equiv 7 \pmod{9}$ since $k^3 \equiv 0,1,8 \pmod{9}$ for all integers $k$. Hence $9 \mid n+2$, which means $n^5 + n^3 + 2n^2 + 2n + 2 \equiv (-2)^5 + (-2)^3 + 2 \cdot (-2)^2 + 2 \cdot (-2) + 2 = -32 - 8 + 8 - 4 + 2 = -34 \equiv 2 \pmod{9}$, implying $n^5 + n^3 + 2n^2 + 2n +$ is not a cube. This contradiction completes the proof.

Solar Plexsus wrote: Assume $2n^2 + n + 2$ is a cube. The fact that $2n^2 + n + 2 \equiv 2,5,3,5,2,3,3,8,3 \pmod{9}$ for $n = 0,1,2, \cdots, 8$, yields $n \equiv 7 \pmod{9}$ since $k^3 \equiv 0,1,8 \pmod{9}$ for all integers $k$. Hence $9 \mid n+2$, which means $n^5 + n^3 + 2n^2 + 2n + 2 \equiv (-2)^5 + (-2)^3 + 2 \cdot (-2)^2 + 2 \cdot (-2) + 2 = -32 - 8 + 8 - 4 + 2 = -34 \equiv 2 \pmod{9}$, implying $n^5 + n^3 + 2n^2 + 2n +$ is not a cube. This contradiction completes the proof. You missed a Case $n\equiv 6\mod 9\implies 2n^2+n+2\equiv -1\mod 9$ So $n\equiv 6\mod 9$ is also possible but then $n^5+n^3+2n^2+2n+2\equiv 5\mod 9$ a contradiction.