I imagine this is probably a duplicate of some other question but I cannot quite find it via the search.
Yes, you can construct a Mordell curve with arbitrarily many integer points if there are no restrictions to $k$. I will rotate the usage of $x$ and $y$ and consider the curve as
$$
E: y^2 = x^3 + k
$$
instead, so we want $y\geq 0$.
The idea is to pick a rank 1 curve, generate $n$ rational points using the generator $G$ and transform them into integer points. Looking at denominators of the points, let $D=LCM(${denominators}$)$. Now multiply by $D^6$ to transform the curve into
$$
E: (D^3y)^2 = (D^2 x)^3 + (D^6k) \Leftrightarrow Y^2 = X^3 + D^6k
$$
Then, by definition of $D$, each of the $n$ points $(D^2x_i,D^3y_i)$ must now be integral.
Example:
Consider the Mordell curve over $\mathbb Q$
$$E: y^2 = x^3 + 2$$
It has rank one and the only torsion point is the neutral point $\mathcal O$. The generator is $G=(-1,1)$, we can find the first 5 rational points as
$$
(x_k,y_k) = [k]G
$$
for $1\leq k\leq 5$, where $[\cdot]$ denotes the point multiplication operation. This gives the following points
$$
\begin{align*}
G &= (-1,1) \\
[2]G &= (17/4,-71/8) \\
[3]G &= (127/441 , 13175/9261) \\
[4]G &= (66113/80656 , -36583777/22906304) \\
[5]G &= (108305279/48846121 , 1226178094681/341385539669)
\end{align*}
$$
Next we find the LCM:
$$
D = LCM(4,8,441,9261,80656,22906304,48846121,341385539669) = 72419917504456587635136
$$
Finally we apply the transformation $(x',y') = (D^2x,D^3y)$:
$$
\begin{align*}
G &= (-5244644451352297667755999615909647085061738496, 379816718507139378146913182774552967922429082711157728996445493395456)\\
[2]G &= (22289738918247265087962998367616000111512388608,
-3370873376750861981053854497124157590311558109061524844843453753884672)\\
[3]G &= (1510362461047033568718847961951304262591475712,
540339624914324728116356892674088689383220296373988022840748231884800)\\
[4]G &= (4298988030800615648040473152730540787228311808,
-606607252341406100174229116883478323397886401063198574917734250057728) \\
[5]G &= (11628818602801899585790299342807326836436987904,
1364214022300501406044776062737876197523992448609719798004173622738944)
\end{align*}
$$
and we check that each $(x,y)$ satisfies the new curve
$$
E': Y^2 = X^3 + 2D^6
$$
