A note on the Diophantine equation $\left( 4^{q}-1\right) ^{u} +\left( 2^{q+1}\right) ^{v}=w^{2}$

Let $a, b$ and $ c $ be positive integers such that $a^{2}+b^{2}=c^{2}$ with $\gcd \left( a,b,c\right) =1$, $a$ even. Terai's conjecture claims that the Diophantine equation $x^{2}+b^{y}=c^{z}$ has only the positive integer solution $(x,y,z)=(a,2,2)$. In this short note, we prove that the equation of the title, has only the positive integer solution $(u,v,w)=(2,2,4^{q}+1),$ where $q$ is a positive integer.