Find $n$ such that $J=n\mathbb Z$

We have $$I=\{(4+3i) (a+bi):a,b\in\mathbb Z\} = \{(4a-3b) + (3a+4b)i:a,b\in\mathbb Z\}.$$ Now you want $I\cap \mathbb Z$, this will be the set of numbers of the form above with imaginary part zero, i.e., integers of the form $4a-3b$ such that $3a+4b=0$, i.e., integers of the form $\frac{25a}{4}$.

It follows that $J=25\mathbb Z$, since $25a/4$ is an integer if and only if $a$ is a multiple of $4$.

Since $J$ is the intersection of an ideal and a subring, it must be an ideal of the subring.

Now, when is an integer divisible by $4+3i$ in the Gaussian integers? To answer this question, suppose that $(4+3i)(a+bi)=m \in \mathbb{Z}$. Then, $4a-3b=m$ and $3a+4b=0$. Multiplying the first equation by $4$ and the second one by $3$, we get $16a-12b=4m$ and $9a+12b=0$. Now, the coefficients of $b$ are opposites, so we can use the elimination method. Adding the two equations gives $25a=4m$. Since $4m$ is divisible by $25$, and $4$ and $25$ are coprime, $m$ must also be divisible by $25$. This proves that $J \subseteq 25\mathbb{Z}$.

Conversely, we also have $25\mathbb{Z} \subseteq J$, since $25=4^2+3^2=(4+3i)(4-3i) \in J$.

Hence, we in fact have that $J=25\mathbb{Z}$, so the two possible values of $n$ that generate the ideal $J$ are $25$ and $-25$.

Since $(4+3i)(a+bi)=(4a-3b) + (3a+4b)i$ we can see that to be in $\mathbb{Z}$ we must have $3a+4b=0$. Since $3$ and $4$ are relatively prime $4|a$ and $3|b$. Choosing $a$ and $b$ as close to $0$ as we can minimal $a$ and $b$ are $a=4$ and $b=-3$ (or $a=-4, b=3$) which together with $4a-3b$ gives us $4^2 + 3^2=25$. Since we chose the smallest $a$ and $b$ that we have, $n=25$, as it is the closest to zero in the intersection.