Since $a^2<a^2+3b<(a+2)^2$, we have $3b=2a+1$. Similarly, $3c=2b+1$, so $a=\frac{9c-5}{4}$. Now $c^2<c^2+3a=c^2+\frac{27}4c-\frac{15}4<(c+4)^2$. Solving the cases $c^2+3a=(c+1)^2,(c+2)^2,(c+3)^2$, we find the solutions $(1,1,1)$ and $(37,25,17)$.

Since $a\ge b$, we have $a^2+3b\le a^2+3a< a^2+4a+4=(a+2)^2$. However, $a^2<a^2+3b$, so $a^2+3b=(a+1)^2\implies 3b=2a+1$. Similarly, $3c=2b+1$. So, $a=\frac{3b-1}{2}$ and $b=\frac{3c-1}{2}$. Thus, $a=\frac{3\cdot \frac{3c-1}{2} - 1}{2} = \frac{9c-5}{4}$. Since $a$ is an integer, $c\equiv 1\pmod 4$, and $c^2+\frac{27c-15}{4}$ is a perfect square. Multiplying both sides by $4$ gives us that $4c^2+27c-15$ is a perfect square. However, $(2c+8)^2=4c^2+32c+64>4c^2+27c-15$ and $(2c)^2<4c^2+27c-15$. Since $4c^2+27c-15$ is even, we have $4c^2+27c-15\in \{(2c+2)^2, (2c+4)^2, (2c+6)^2\}$ Case 1: $4c^2+27c-15=(2c+2)^2$. Then $4c^2+27c-15=4c^2+8c+4\implies 19c=19\implies c=1$. So we get the solution $\boxed{(a,b,c)=(1,1,1)}$, which works. Case 2: $4c^2+27c-15=(2c+4)^2$. Then $4c^2+27c-15=4c^2+16c+16\implies 11c=31$, so $c$ is not an integer, contradiction. Case 3: $4c^2+27c-15=(2c+6)^2$. Then $4c^2+27c-15=4c^2+24c+36\implies 3c=51\implies c=17$. So we get $\boxed{(a,b,c)=(37, 25, 17)}$, which works.

Here's a link to a video solution https://youtu.be/AzO0UdNVHuQ