The diagonal elements just shift the spectrum (and the top eigenvalue) by $1$. So we may assume they are $0$.
All depends on what $k$ is. If $k/n^2\to \alpha\in (0,1/2)$, you are essentially dealing with a symmetric matrix whose entries above the diagonal are iid Bernoulli $\{0,1\}$ of parameter (=mean) $p=2\alpha$ and variance $\sigma^2=p(1-p)$. (I will discuss below the difference between your model and the Bernoulli one). Let $X$ be this Bernoulli matrix. Write $X=\sqrt{n} (Y+p/\sqrt{n} 1 1^T)=\sqrt{n} Z$ where the $Y$'s now have mean $0$ and variance $1/n$. It is a standard fact that there is a transition: if $p\leq \sigma$ then the largest eigenvalue of $Z$ concentrates at $2\sigma$. If $p>\sigma$ then it concentrates at $p+\sigma^2/p$. Fluctuation results are also known, see https://arxiv.org/pdf/math/0605624v1.pdf
Note that your model is sandwiched whp between $p_-=p-K/\sqrt{n}$ and $p_+=p+K/\sqrt{n}$, so your model behaves the same (up to multiplication by $\sqrt{n}$ and shift by $1$).
If on the other hand $k/n^2\to 0$, you have that the rank 1 perturbation disappears (formally, note that this is the behavior when $p\to 0$ and $p/\sigma^2\to 1$, so $p<\sigma$). But in this case, of course the scaling is not $\sqrt{n}$ but rather $\sqrt{n}\sigma \sim \sqrt{pn}$. This will work as long as $k$ is not too small, I have not checked exactly the threshold, but $k/n\to \infty$ faster than some poly log should work https://arxiv.org/abs/1309.4922
Finally, if $k/n\to 0$, you will have many empty rows and columns, so you need to first erase those. I did not check and am not sure what happens there.