Let $v(v+3) = w^2$, $w \in \mathbb{Z}$. We assume that: $v = 3\kappa$, $\kappa \in \mathbb{N}^*$. Then we have:
$$v(v+3) = 3\kappa(3\kappa+3) = 9\kappa(\kappa+1) = w^2 \Leftrightarrow 3 \cdot 3\kappa(\kappa+1) = w^2 \Rightarrow 3|w^2$$
from which, since 3 is a prime, it follows that $3|w$.
$$\text{Hence } \kappa(\kappa+1) = \left(\frac{w}{3}\right)^2, \text{ with } \frac{w}{3} \in \mathbb{Z}, \text{ and thus we have}$$
$$\kappa^2 < \kappa(\kappa+1) = \left(\frac{w}{3}\right)^2 < (\kappa+1)^2 \Rightarrow \kappa < \left|\frac{w}{3}\right| < \kappa+1, \text{ a contradiction.}$$
Hence $v$ is not a multiple of 3.

Alternative solution:
Let $v(v+3) = w^2$, $w \in \mathbb{Z}$. Then we have
$$\begin{align*} v^2 + 2v + 1 \le v^2 + 3v < v^2 + 4v + 4 &\Rightarrow (v+1)^2 \le w^2 < (v+2)^2, w \in \mathbb{Z} \\ &\Rightarrow w = v+1 \text{ if } w = -(v+1). \end{align*}$$
It follows that
$$v(v+3) = (v+1)^2 \Leftrightarrow 3v = 2v+1 \Leftrightarrow v = 1 \ne \pi o \lambda .3.$$