Solve $(x+1)(y+1)(z+1)=144$ in primes "Solve $(x+1)(y+1)(z+1)=144$ in primes".
So far, I have concluded that the solutions are $(x,y,z)=(2,3,11)$ or $(2,5,7)$ and their permutations. I worked like this:


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*$x \equiv 0\mod 2\Rightarrow x+1=3, 144=2^4*3^2 \Leftrightarrow (y+1)(z+1)=48=2^4*3$


*

*$y \equiv 0\mod 2\Rightarrow y+1=3 \Leftrightarrow z=15$, contradiction since $15$ is not prime

*$y \equiv 1\mod 2\Rightarrow y+1\geq 2^2 \Leftrightarrow y+1=2*3, z+1=2^3$ or $y+1=2^2*3, z+1=2^2 \Leftrightarrow y=5, z=7$ or $y=11, z=3$
And since the equation is symmetric the solutions are the permutations of the latter ones. Similarly, through casework, we find the same solutions if $x \equiv 1\mod 2$ (if I haven't made a mistake). My question is if there exists a more simple way to solve this problem besides lots of casework (and if there exists another triplet that satisfies the given equation).

Denote $x' := x + 1$ and analogously. As you point out, solutions are closed under permutation, so we may suppose w.l.o.g. that $x' \leq y' \leq z'$.
Since the factors $x', y', z'$ are all at least $3$, we must have $z + 1 \leq \frac{144}{3^2} = 16$. But the factors of $144 = 2^4 \cdot 3^2$ no larger than $16$ are
$1, 2, 3, 4, 6, 8, 9, 12, 16$, and the numbers among these $1$ larger than a prime are $\{3, 4, 6, 8, 12\}$. We can reduce the number of cases to check by observing that $5^3 < 144 < 6^3$, which implies that $x' \in \{3, 4\}$ and $z' \in \{6, 8, 12\}$.


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*If $z' = 6$, then $x' y' = \frac{144}{6} = 24$ and $\sqrt{24} \leq y' \leq 6$, so $y' = 6$: $(x', y', z') = (4, 6, 6)$.

*If $z' = 8$, then $x' y' = \frac{144}{8} = 18$ and $\sqrt{18} \leq y' \leq 8$, but the only factor of $18$ in that range is $6$: $(x', y', z') = (3, 6, 8)$.

*If $z' = 12$, then $x' y' = \frac{144}{12} = 12$, so $x' \leq \sqrt{12}$ and thus $x' = 3$: $(x', y', z') = (3, 4, 12)$.


The solutions are: $$(x, y, z) = (2, 3, 11), (2, 5, 7), (3, 5, 5) .$$ 
It seems the error in the original post was in the unwritten details of the case $x \equiv 1 \pmod 2$.
