This is not a complete solution, but it may help. Suppose the equation has a solution. Then use the Wicks forms (Lyndon, R. C.; Wicks, M. J. Commutators in free groups. Canad. Math. Bull. 24 (1981), no. 1, 101–106.): there exist reduced decompositions $u\equiv a\bar c, x\equiv cb$, $v\equiv p\bar q, u\equiv qr$ such that $[u,x]\equiv abc\bar a\bar b\bar c, [v,u]\equiv pqr\bar p\bar q\bar r$ where $\equiv$ means there are no cancellations in these words, and $\bar z=z^{-1}$. Since $abc\bar a\bar b\bar c=pqr\bar p\bar q\bar r$, we have that either $\bar c$ is a suffix of $\bar r$ or vice versa, etc. The point is that with Wicks forms you can work as in the free semigroup (no cancellations).
This is not a complete solution, but it may help. Suppose the equation has a solution. Then use the Wicks forms (Lyndon, R. C.; Wicks, M. J. Commutators in free groups. Canad. Math. Bull. 24 (1981), no. 1, 101–106.): there exist reduced decompositions $u\equiv a\bar c, x\equiv cb$, $v\equiv p\bar q, u\equiv qr$ such that $[u,x]\equiv abc\bar a\bar b\bar c, [v,u]\equiv pqr\bar p\bar q\bar r$ where $\equiv$ means there are no cancellations in these words, and $\bar z=z^{-1}$. Since $abc\bar a\bar b\bar c=pqr\bar p\bar q\bar r$, we have that either $\bar c$ is a suffix of $\bar r$ or vice versa, etc.