Nice problem! Will edit this, but solution sketch: Let $x,y >0$ be positive integer solutions to \[ 2^nx - 3^ny = 1\]We know by Bézout, there exists such a solution with $0 < x' < 3^n$ and $0<y'<2^n$. We will show that we can find a solution with $gcd(2,x) = gcd(3,y)= 1$. If $x'$ is odd, then one of $y'$ or $y' + 2^{n+1}$ isn't divisible by $3$, meaning one of $(x',y')$ or $(x' + 2 \cdot 3^n, y' + 2^{n+1}$) suffices the wanted gcd condition. If $x'$ is even, then one of $y' + 2^n$ or $y' + 3 \cdot 2^n$ isn't divisible by $3$, meaning one of $(x' + 3^n, y' + 2^n)$ or $(x' + 3^{n+1}, y' + 3 \cdot 2^n)$ suffices the wanted gcd condition. Notice that in both cases we found a solution with $x < 3^{n+1} + 3^n$ and $y < 2^{n+2}$, and we have $x,y < 5^n$ for large enough $n$ (say $n > N$). This combined with $gcd(6,x) = gcd(6,y) = 1$ we have $f(x) < n$ and $f(y) < n$ for large enough $n$ (notice that $gcd(3,x) = gcd(2,y) = 1$ is obvious). If we now take $m + 1= x \cdot 2^n$ and $m = y \cdot 3^n$, and take $n>N$, then these infinite $m$ will always work. Maybe incorrect right now, but idea to finish off; More generally, we can show that the equation \[x_u \cdot p_u^n - x_{u+1} \cdot p_{u+1}^n = -1\]has a solution with $gcd(\prod_{i=1}^{u+1}p_i, x_u) = gcd(\prod_{i=1}^{u+1}p_i, x_{u+1}) = 1$, implying $x_u, x_{u+1} < p_{u+2}^n$ (where we set $n$ very large) and thus $f(x_u),f(x_{u+1})<n$, meaning setting $m+u = x_up_u^n$ will work, for all $0 \le u \le k$

Fix a large integers $N, T$ and let $p_i$ be the $i$th prime. Let $P = (p_{T+1}p_{T+2} \dots p_{T+k})^N$. Then let $(a_1, \dots, a_k)$ be tuples with $a_i < P$ and $a_{i} p_{T+i}^k$ increases by $1$ when $i$ increases by $1$ for each $i$ by pigeonhole. Consider tuples of the form \[ \left(\left(a_1 + C \cdot \frac{P}{p_{T+1}^N}\right) p_{T+1}^N, \left(a_2 + C \cdot \frac{P}{p_{T+2}^N}\right) p_{T+2}^N, \dots, \left(a_k + C \cdot \frac{P}{p_{T+k}^N}\right) p_{T+k}^N \right) \]We know that $a_i + C \cdot \frac{P}{p_{T+i}^N}$ for $C \in [0, x]$ is divisible by $p^{N+1}$ or $p_{T+i}$ for some prime $p$ for at most $\left(\sum_{i=1}^\infty \frac{1}{p_i^N} + \sum_{i=1}^k \frac{1}{p_{T+i}}\right) $ which is less than $\frac{1}{k}$ for large $T$ and $N$. Summing over $i$ means that there exists a $C$ for which \[ f\left(\left(a_i + C \cdot \frac{P}{p_{T+i}^N}\right) p_{T+i}^N\right) = N \]for each $i = 1, 2, \dots, k$ as desired.