No. Suppose the contrary, and let $a,b\in\mathbb{N}$ be side lengths of such a (right-angled) triangle. Then we have \[ a^2+b^2 = 2^{5\cdot 4034} \cdot 3^{2\cdot 4034}\cdot 7^{4034}. \]Now, if $7\nmid a$ then $7\nmid b$, but in this case, $(a\cdot b^{-1})^2\equiv -1\pmod{7}$, a contradiction. Hence, $7\mid a,b$. Applying this idea repeatedly, we find $7^{2017}\mid\mid a,b$ and likewise $3^{4034}\mid\mid a,b$. With this, it suffices to show there are no $m,n\in\mathbb{N}$ such that \[ m^2+n^2 = 2^u,\qquad\text{where}\qquad u\equiv 0\pmod{2} \]and $u>2$. Clearly $m,n$ are even, set $m=2m_1$ and $n=2n_1$, and obtain $m_1^2+n_1^2 = 2^{u-2}$. Continuing in this manner, we finally are left with $m_t^2+n_t^2=2^2$ for some $t\in\mathbb{N}$, but this is clearly not possible.