Let $v^2 + 10v + 160 = \kappa^2$, where $\kappa$ is a positive integer. Then we have:
$$\begin{align*} v^2 + 10v + 160 = \kappa^2 &\Leftrightarrow (v+5)^2 + 135 = \kappa^2 \Leftrightarrow \kappa^2 - (v+5)^2 = 135 \\ &\Leftrightarrow (\kappa - |v+5|)(\kappa + |v+5|) = 135 = 1 \cdot 3^3 \cdot 5, \text{ where } \kappa > |v+5| \\ &\Leftrightarrow \begin{cases} \kappa - |v+5| = 1 \\ \kappa + |v+5| = 135 \end{cases} \text{ or } \begin{cases} \kappa - |v+5| = 3 \\ \kappa + |v+5| = 45 \end{cases} \\ &\text{or } \begin{cases} \kappa - |v+5| = 5 \\ \kappa + |v+5| = 27 \end{cases} \text{ or } \begin{cases} \kappa - |v+5| = 9 \\ \kappa + |v+5| = 15 \end{cases} \\ &\Leftrightarrow \begin{cases} \kappa = 68 \\ |v+5| = 67 \end{cases} \text{ or } \begin{cases} \kappa = 24 \\ |v+5| = 21 \end{cases} \text{ or } \begin{cases} \kappa = 16 \\ |v+5| = 11 \end{cases} \text{ or } \begin{cases} \kappa = 12 \\ |v+5| = 3 \end{cases} \end{align*}$$

$$\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l} \kappa = 68 \\ v + 5 = \pm 67 \end{array} \right\} \text{ or } \left\{ \begin{array}{l} \kappa = 24 \\ v + 5 = \pm 21 \end{array} \right\} \text{ or } \left\{ \begin{array}{l} \kappa = 16 \\ v + 5 = \pm 11 \end{array} \right\} \text{ or } \left\{ \begin{array}{l} \kappa = 12 \\ v + 5 = \pm 3 \end{array} \right\} \\[1.5ex] \Leftrightarrow \left\{ \begin{array}{l} \kappa = 68 \\ v = 62 \ \text{or} \ v = -72 \end{array} \right\} \text{ or } \left\{ \begin{array}{l} \kappa = 24 \\ v = 16 \ \text{or} \ v = -26 \end{array} \right\} \\[1.5ex] \text{or } \left\{ \begin{array}{l} \kappa = 16 \\ v = 6 \ \text{or} \ v = -16 \end{array} \right\} \text{ or } \left\{ \begin{array}{l} \kappa = 12 \\ v = -2 \ \text{or} \ v = -8 \end{array} \right\}. \end{array}$$
Therefore $v \in \{-72, -26, -16, -8, -2, 6, 16, 62\}$.