It is easy to see a solution $(10,1,1)$.Consider another solution $(x,y,z)$.We can take $mod 5$ and we find that $x=4k+2$.Also $y$ is odd $mod 4$ and $z$ is odd $mod 8$. We have $2^{10}(16^{k-2}-1)=15(3^{y-1}5^{z-1}-1).y-1=2q,z-1=2f$. Assume q>0. mod 7 we have $16^{k-2}-1=0(mod 7)$ and we have 4(k-2)=0(mod 3) so $ v_3(16^{k-2}-1)=v_3(15)+v_3(k-2) $so $y=1$.Assume $f>0$. We have now the equation $2^{10}(16^{k-2}-1)=15(5^{2f}-1)$ and we take LTE for 3.We have $ v_3(16^{k-2}-1)=1+v_3(k-2)$ but $5^{2f}-1=0(mod 3)$ so we have that $k-2=0(3)$ so $16^{k-2}-1=0(7)$ so f=0(3)(because order)So $5^2f-1=0(31)$ so $16^{k-2}-1=0(31)$ so $k-2=0(5)$ and now LTE $v_5(16^{k-2}-1) \ge 2$ which is a contradicition so q=0.So the only solution $(10,1,1)$

Could this be solved without Lifting the exponent lemma?

@above Assume that $y\geq2$ We have $2^x \equiv 1 \pmod{9}$, which implies $x \equiv 0 \pmod6$. It follows that $2^x-1009\equiv 0 \pmod{7}$, which is impossible. Assume that $z\geq2$ We have $2^x \equiv 9 \pmod{25}$, which implies $x \equiv 14 \pmod{20}$. It follows that $2^x-1009\equiv 0 \pmod{41}$, which is impossible. Thus we must have $y=z=1$. Therefore, $(x,y,z)=(10,1,1)$ is the only solution.

Thank you very much!