Consider how you would generate some natural numbers with this method. I'll simplify your notation by your sums as multiples of $3$ and $5$.
$$\begin{aligned}\color{red}{1}\times3+\color{blue}{1}\times5&=8\\
\color{red}{3}\times3+\color{blue}{0}\times5&=9\\
\color{red}{0}\times3+\color{blue}{2}\times5&=10\\
\color{red}{2}\times3+\color{blue}{1}\times5&=11\\
\ldots
\end{aligned}$$
Can you notice any patterns appearing? Consider how you change the red and blue multiples of $3$ and $5$, if you wanted to increase the right-hand-side by $1$.
$$\begin{aligned}&\color{red}{1}&&\color{blue}{1}&&8\\
&\color{red}{\downarrow(+2)}&&\color{blue}{\downarrow(-1)}&&\downarrow(+1)\\
&\color{red}{3}&&\color{blue}{0}&&9\\
&\color{red}{\downarrow(-3)}&&\color{blue}{\downarrow(+2)}&&\downarrow(+1)\\
&\color{red}{0}&&\color{blue}{2}&&10\\
&\color{red}{\downarrow(+2)}&&\color{blue}{\downarrow(-1)}&&\downarrow(+1)\\
&\color{red}{2}&&\color{blue}{1}&&11\\
\end{aligned}$$
You should be able to see that we can increase the right-hand-side by $1$ by either:
We can continue in this regard to generate more integers, using whichever of these two rules that lets us keep a nonzero number of $3$'s and $5$'s. But why does this work? Fundamentally, it is because $\color{red}{2}\times3+\color{blue}{(-1)}\times5=1$. But also, $\color{red}{(-3)}\times3+\color{blue}{2}\times5=1$. Hence, we can raise our right-hand-side by $1$ by changing the number of $3$'s and $5$'s on left-hand-side.
If there are integers $x,y$, such that the integers $a,b$ satisfy $xa-yb=1$, then $a$ and $b$ are coprime. Hence, coprime natural numbers are the numbers you're looking for. An easy way to generate an integer $b$, that is coprime to an integer $a$ is to pick a $b$ that shares no factors with $a$. For example if $a=3$, you could pick $4,5,7,8,10,\ldots$.
However I'm not quite sure how to prove that you will always have enough $3$'s and $5$'s to allow you to generate arbitrarily large numbers.