This is a good question in 2023 Taiwan TST Round 3, I think solving it can improve skills for solving inequalities and Diophantine equations . Here is my solution: Suppose that there are positive integers $k_{1}, k_{2}$ such that $ab = k_{1}^2, c = \frac{k_{2} ^ 3}{k_{1} ^ 2}$. Then we have $3k_{2} + 1 = a + b + c \geq 2k_{1} + \frac{k_{2} ^ 3}{k_{1} ^ 2} \geq 3k_{2}$, and $2k_{1} + \frac{k_{2} ^ 3}{k_{1} ^ 2} = 2k_{1} + c$ is a postive integer. Case 1: $2k_{1} + \frac{k_{2} ^ 3}{k_{1} ^ 2} = 3k_{2} + 1$ Now we let $k_{2} = k_{1} + d$, then we have $3d ^ {2} k_{1} + d ^ 3 = k_{1} ^ 2$, i.e. $d ^ 2(d + 3k_{1}) = k_{1} ^ 2$. Suppose that $h = \frac{k_{1}}{d}$, then $h^2 = (3h + 1)d$, there is no solutions of $h, d$ are integers. Case 2: $2k_{1} + \frac{k_{2} ^ 3}{k_{1} ^ 2} = 3k_{2}$ Now we have$ k_{2} = k_{1}$, and we let that equals $k$. So $ab = k^2, c = k$, and the equation $a + b + k - 3k = 1$ holds, so that means $a + b = 2k + 1$. We let $d = k - a$, then $a = k - d, b = k + d + 1$ implies $k = d(d+1)$, i.e. there is only a solution $(a, b, c) = (d ^ 2, d ^ 2 + 2d + 1, d ^ 2 + d)$. From the above two cases, we always have only a solution $(a, b, c) = (d ^ 2, d ^ 2 + 2d + 1, d ^ 2 + d) \forall d \in \mathbb{N}$.

Slightly different writeup of the same idea: By AM-GM, we have \[a+b+c \ge 2\sqrt{ab}+c \ge 3\sqrt[3]{abc}\]and all three are integers with the RHS and LHS differing by exactly $1$. Hence either $a=b$ or $c^2=ab$. In the first case, we have $a^2c$ a perfect cube and $(a;c)=1$ so that $a=x^3, c=y^3$ and we get \[1=2x^3+y^3-3x^2y=(x-y)^2(2x+y)\]which is clearly impossible. In the second case, we have $a+b=2c+1$, hence $(\sqrt{a}-\sqrt{b})^2=1$ and hence $\sqrt{a}-\sqrt{b}=\pm 1$ so that both $a,b$ are perfect squares and we obtain the family of solutions $(k^2,(k+1)^2,k(k+1))$ and $((k+1)^2,k^2,k(k+1))$ for $k \in \mathbb{N}$.

@above Slight typo, I think you meant \[ a + b + c \ge 2 \sqrt{ab} + c \ge 3 \sqrt[3]{abc} \]