let $X_1, \dots, X_n$ be a random sample from a binomial $\text{bin}(k,p)$ population where $p$ is known and $k$ is unknown. We attempt to maximize $L(k | x,p)$ the likelihood function without differentiation.
If $k \lt \text{max}_i x_i$ then $L(k | x, p) = 0$, so the MLE is an integer $k \ge \text{max}_i x_i$ that satisfies $L(k|x,p) / L(k-1 | x, p) \ge 1$ and $L(k+1 | x, p) / L(k | x,p) \lt 1$.
Why does $k$ have to satisfy these two inequalities? And how can it satisfy the second inequality if it satisfies the first?
Since $L(k | x,p) = \prod_{i=1}^n $ $k\choose x_i$ $p^{x_i}(1-p)^{k-x_i}$, if the first inequality is true then the second inequality should be false.
If you're trying to find the maximum of $L(k\mid x,p)$ for any $k$ then you should be finding a $k$ such that $L(k \mid x,p) \geq L(m \mid x,p)$ for any $m.$
In particular, setting $m = k - 1,$ you would have $$ L(k \mid x,p) \geq L(k - 1 \mid x,p) \tag1 $$ and setting $m = k + 1$ you would have $$ L(k \mid x,p) \geq L(k + 1 \mid x,p). \tag2$$ But $(1)$ gives you $$ \frac{ L(k \mid x,p)}{L(k - 1 \mid x,p)} \geq 1, $$ which is one of the inequalities you're supposed to have, and $(2)$ gives you $$ \frac{ L(k + 1\mid x,p)}{L(k \mid x,p)} \leq 1 , $$ which is almost the other inequality; the only change is that you were supposed to have $<$ instead of $\leq.$ But unless the values of $L(m \mid x,p)$ for $m > k$ are all equal to each other then at some point you will have $<$ instead of $\leq,$ so choose your $k$ to be on the left side of the first $<$ inequality.