Non primality of a pseudo-like Mersenne numbers [closed]

Are there infinitely many $$m\in\mathbb N$$ such that $$35\times2^m+1$$ is $$\textbf{not}$$ a prime number?

• If $m$ is even the number is divisible by $3$. – Dag Oskar Madsen Oct 13 at 9:10
• In general, it is easy to see that for any $a,b,c\in\mathbb Z$ such that $a\ne0$ and $b\ne0,\pm1$, there is an infinite arithmetic progression of $m$’s such that $ab^m+c$ is composite. – Emil Jeřábek Oct 13 at 9:51
• Any reason for $35$ in particular? – Wojowu Oct 13 at 10:17
For natural $$k$$ let $$m=4+10k$$. Then $$35 \times 2^m+1$$ is divisible by $$11$$, since $$35 \cdot 2^{4+10k} \equiv -1 \pmod {11}$$ and $$2^m$$ is periodic modulo all primes.
For any positive integer $$m$$, there are infinitely many $$n$$ such that $$35 \times 2^n+1$$ is divisible by $$35 \times 2^m+1$$. Indeed, it suffices to take $$n$$ such that $$2^{n-m} \equiv 1 \mod 35 \times 2^m+1$$.