Solution:

Since the integer bases are in ratio $8 : 7,$ they are $8a$ and $7a$ for a positive integer $a.$ Equal areas make the corresponding altitudes inversely proportional to the bases, say $7h$ and $8h.$ The legs are then $\sqrt{16a^2 + 49h^2}$ and $\sqrt{\frac{49}{4}a^2 + 64h^2},$ and equal perimeters give $$8a + 2\sqrt{16a^2 + 49h^2} = 7a + 2\sqrt{\tfrac{49}{4}a^2 + 64h^2}.$$

Moving $7a$ to the left and squaring yields $a\sqrt{16a^2 + 49h^2} = 15h^2 - 4a^2;$ squaring again and simplifying leaves $225h^4 = 169a^2h^2,$ so $h = \frac{13a}{15}.$ The legs become $$\sqrt{16a^2 + 49 \cdot \tfrac{169a^2}{225}} = \frac{109a}{15} \qquad \text{and} \qquad \sqrt{\tfrac{49}{4}a^2 + 64 \cdot \tfrac{169a^2}{225}} = \frac{233a}{30}.$$

For all sides to be integers, $30$ must divide $a.$ Taking $a = 30$ gives the triangles $(218, 218, 240)$ and $(233, 233, 210),$ each with perimeter $676$ and area $21840.$ The minimum common perimeter is $676.$