If we multiply through by $pab$, we find that for $a, b \ne 0$, our equation is equivalent to $pb+pa=ab$, which we can rewrite as
$$(a-p)(b-p)=p^2.$$
We are looking for non-zero solutions of this equation. Conveniently, $p^2$ does not have many factorizations!
We can have $a-p=-1$, $b-p=-p^2$, which yields a negative $b$, or $a-p=-p^2$, $b-p=-1$, which yields a negative $a$.
We could have $a-p=-p$, $b-p=-p$, but that yields the impossible $a=b=0$.
We can have $a-p=1$, $b-p=p^2$, or $a-p=p^2$, $b-p=1$, which respectively yield the solutions $a=p+1$, $b=p^2+p$, and $a=p^2+p$, $b=p+1$.
Finally, we can have $a-p=p$, $b-p=p$, which yields $a=b=2p$.
Comment: The same idea can be used to find all integer solutions of the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$, where $n$ is a given positive integer. The factorizations of $n^2$, with the exception of $n^2=(-n)(-n)$, supply the integer solutions.
Define the Fibonacci sequence in the usual way: $F_0=0$, $F_1=1$, $F_{k+2}=F_{k+1}+F_k$ for $k\ge0$. It is easy to prove by induction that for all indices $k$ we have
$$
F_{k+1}^2-F_{k+1}F_k-F_k^2=(-1)^k.\tag{1}
$$
Assume that $k$ is odd. Substitute $F_{k+1}=F_{k+2}-F_k$ to equation $(1)$. After expanding it out and combining the terms we get
$$
F_{k+2}^2-3F_{k+2}F_k+F_k^2=-1.\tag{2}
$$
Let us select $m=F_{k+2}+1$ and $n=F_k+1$. Plugging these into equation $(2)$ gives
$$
(m-1)^2-3(m-1)(n-1)+(n-1)^2=-1\Leftrightarrow (m^2+m)+(n^2+n)=3mn,
$$
meaning that the choices $m=F_{k+2}+1$, $n=F_k+1$ is a solution to the equation
$$
\frac{m+1}n+\frac{n+1}m=3.
$$
Obviously the infinitude of the number of solutions follows from this.
How did I come up with this? Crunch out a few thousand test cases by Mathematica. Observe that $3$ occurs as the sum often. Solve for $m$ in terms of $n$ from the equation
$$
\frac{m+1}n+\frac{n+1}m=3.
$$
It turns out that the discriminant of this quadratic equation is
$$
5n^2-10n+1=5(n-1)^2-4.
$$
From pleasant pieces of personal history (IMO1981) I knew that this is a perfect square, if
$n-1$ is a Fibonacci number of an odd index.
Best Answer
You are looking for coprime integers $a$ and $b$ for which $$\frac{a}{b}+14\frac{b}{a}=\frac{a^2+14b^2}{ab},$$ is an integer. Then $b$ divides $a^2$, and so $b=1$ because $\gcd(a,b)=1$. Now the above reduces to $$\frac{a^2+14}{a},$$ which is an integer if and only if $a$ divides $14$, i.e. $a\in\{1,2,7,14\}$.