Let $M$ be the midpoint of $BC:$ right triangles at $M$ give $\cos\angle ABM = \frac{BM}{AB}$ and $\cos\angle IBM = \frac{BM}{8}$

Since $\angle IBM$ is half of $\angle ABM,$ the double-angle formula relates $AB$ to $BC;$ integrality and $BM \lt 8$ leave only a few cases to test

Let $M$ be the midpoint of $\overline{BC};$ by symmetry $A,$ $I,$ and $M$ are collinear with $AM \perp BC.$ With $a = AB$ and $b = BM,$ right triangles $ABM$ and $IBM$ give $\cos\angle ABM = \frac{b}{a}$ and $\cos\angle IBM = \frac{b}{8}.$ Since $BI$ bisects $\angle ABM,$ the double-angle formula yields $$\frac{b}{a} = 2\left(\frac{b}{8}\right)^2 - 1,$$ so $$a = \frac{32b}{b^2 - 32}.$$

Writing $c = BC = 2b,$ this becomes $a = \frac{64c}{c^2 - 128}.$ We need $c^2 \gt 128,$ so $c \ge 12,$ while $\cos\angle IBM = \frac{b}{8} \lt 1$ forces $c \lt 16.$ Testing $c = 12, 13, 14, 15,$ only $c = 12$ makes $a$ an integer, namely $a = \frac{768}{16} = 48.$

The triangle with sides $48,$ $48,$ $12$ satisfies all the conditions, and its perimeter is $48 + 48 + 12 = 108.$