Assume one of them is $1$, say $c = 1$. Then, $\frac{(b+1)^4}{a} + \frac{(a+1)^4}{b}$ must be an integer. This means $a \mid b(b+1)^4$. Since $a+b+1$ is prime, $\text{gcd}(a,b+1)=1$. And so $a \mid b$, similarly $b \mid a$, implying $a=b$. This gives $\frac{2(a+1)^4}{a}$ is an integer, so $a \mid 2$, giving us the solutions $(1,1,1)$ and $(2,2,1)$. (Also the permutations) Now assume $a,b,c > 1$. Note that $a \mid bc(b+c)^4 \implies a \mid bc$, and similarly $b \mid ac , c \mid ab$. Let $P_{a,b} , P_{a,c} , P_{b,c}$ denote the set of primes that divide both $a,b$ , both $a,c$ and both $b,c$ respectively. Since their sum is a prime, these sets are disjoint. Let $p \in P_{a,b}$ be an arbitrary prime. Then $a \mid bc \implies v_p(a) \leq v_p(b)$ and $b \mid ac \implies v_p(b) \leq v_p(a)$. So $v_p(a) = v_p(b)$. The same is true for other sets. Thus we can write $a = xy , b = xz , c = yz$ with $1 = \text{gcd}(x,y) = \text{gcd}(x,z) = \text{gcd}(y,z)$ and that at most one of $x,y,z$ is $1$. Also we might assume without loss of generality that $x > y > z$. Now the first condition gives $x \mid y+z , y \mid x+z , z \mid x+y$. Since $y + z < 2x$, this gives $y+z = x$. So $y \mid x+z = 2x - y \implies y \mid 2x \implies y \mid 2 \implies y = 2$ and $z = 1$. So $x = 3$ and these give us the solution $(6,3,2)$. It is clear that this is also a solution. So all solutions are $(1,1,1) , (2,2,1) , (6,3,2)$. (upto permutations)