About the power subgroups of the Bianchi groups Let $B$ be the  Bianchi group and let $B^2 =\langle x^2 | x \in B\rangle$, the power subgroup of $B$. 

Is it true that $B \ne B^2$ all the time ?  

 A: To answer the question more concretely: Except for $PSL(2,\mathcal{O}_3)$, yes, every Bianchi group has the property that $H_1(B,\mathbb{Z}/2\mathbb{Z}) \ne 1$. Furthermore, some stronger statements than those covered below are discussed here: Torsion in cuspidal cohomology.
First, let's deal with the exceptional case. $PSL(2,\mathcal{O}_3) \cong \langle a,b,c | a^3, b^3, c^2, (bc)^3, (ca)^3, (ab)^3\rangle$. $H_1(PSL(2,\mathcal{O}_3),\mathbb{Z}) \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$. In terms of intuition, $\mathbb{H}^3/PSL(2,\mathcal{O}_3)$ is a tetrahedral orbifold with a $S^2(3,3,3)$ cusp. (Note $H_1(PGL(2,\mathcal{O}_3), \mathbb{Z}/2\mathbb{Z})=\mathbb{Z}/2\mathbb{Z}.$) For more background Chapters 4.7 and 13 of see Maclachlan and Reid's book The Arithmetic of Hyperbolic 3-Manifolds.
The other case that requires special attention is $PSL(2,\mathcal{O}_1)$. Here $PSL(2,\mathcal{O}_1) = \langle x,y,z,w| x^2,y^2,w^2,(zx)^2,(zy)^2,(zw^2), (yx)^2,(wx)^3 \rangle. $ $H_1(PSL(2,\mathcal{O}_1),\mathbb{Z})= \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ (see page 58 of The Arithmetic of Hyperbolic 3-Manifolds). This is an orbifold with a pillow case cusp. 
In all other cases, we have that $d\ne 1,3$. Here is a simple argument that in these cases,  $\mathbb{H}^3/PSL(2,\mathcal{O}_d)$ has a torus cusp at $\infty$, and so by the "half lives half dies" theorem $\mathbb{H}^3/PSL(2,\mathcal{O}_d)$ will have a non-trivial  $\mathbb{Z}/2\mathbb{Z}$ homology. 
The peripheral subgroup at $\infty$ has a $\mathbb{Z}\times \mathbb{Z}$ subgroup generated by $\pmatrix{1&1\\0 & 1}$ and $\pmatrix{1&\omega\\0 & 1}$ where $\omega = \frac{1+\sqrt{-d}}{2}$ if $d \equiv 3$ mod 4, $\omega = 1+\sqrt{-d}$, otherwise. In either case, the peripheral elements at $\infty$ are of the form $\pmatrix{x & y \\0 & x^{-1}}$, with $x$ a unit in $\mathcal{O}_d$. However, the only units are $\pm 1$ and so $\pmatrix{x & y \\0 & x^{-1}}$ is not a torsion element. Thus, we have established the claim that $\mathbb{H}^3/PSL(2,\mathcal{O}_d)$ has a torus cusp at $\infty$.
