Let $a^2+b^2+c^2=n$ and $n|a^3b+1,b^3c+1,c^3a+1$ obvious, that $(n,a)=(n,b)=(n,c)=1$ $a=b=c=1$ is not solution $n| a^3b-b^3c=b(a^3-b^2c) \to n |a^3-b^2c,b^3-c^2a, c^3-a^2b$ $a^3-b^2c \equiv b^3-c^2a \pmod{n} \to a(a^2+c^2) \equiv b^2(b+c) \pmod{n} \to b^2(b+c)+ab^2 \equiv 0 \pmod{n} \to a+b+c \equiv 0 \pmod {n}$ But $a+b+c < a^2+b^2+c^2$ so contradiction Or in one line: $ n| c( a^3b-b^3c)-b(b^3c-c^3a)= abc(a^2+b^2+c^2)-b^3c(a+b+c)) \to n| a+b+c$ which is impossible

Assume such $a,b,c$ exist. The condition implies that $$a^2+b^2+c^2\mid\sum_{cyc}c(a^3b+1)=abc(a^2+b^2+c^2)+a+b+c.$$Hence $a^2+b^2+c^2\mid a+b+c$. This forces $a=b=c=1$, which is not a solution. Hence no such $a,b,c$ exist.

Suppose all three numbers are divisible by $a^2+b^2+c^2=m$. Obviuosly, $gcd(m,a)=gcd(m,b)=gcd(m,c)=1$ $a^3b-b^3c \vdots m$, so $a^3 \equiv b^2c$. Similarly we get two other cyclic equalities Now $a(-b^2-c^2) \equiv a^3 \equiv b^2c$, from which we get $b^2c+ab^2+ac^2 \equiv b^2c+ab^2+b^3=b^2(a+b+c) \vdots m$ Hence $m=a^2+b^2+c^2>a+b+c \vdots m$. One can easily check $a=b=c=1$ doesn't work

Assume not. Then $$a^2+b^2+c^2|c(a^3b+1)+a(b^3c+1)+b(c^3a+1)=abc(a^2+b^2+c^2)+a+b+c\Rightarrow a^2+b^2+c^2|a+b+c\Rightarrow (a,b,c)=(1,1,1),$$a contradiction.