Problem Statement.
Let $n,p$ be positive integers where $p$ is a prime number and $n<p$. Consider the following linear Diophantine equation:
$$xn-yp=1\tag{1}$$
I am trying to prove that there exists positive integer-pair solutions $(x,y)$ with $x,y<p$.
My Attempted Proof. [Flawed, See Below]
Since $p$ is prime, $\gcd (n,p)=1$. Therefore, by Bézout's identity, eq. (1) always has at least one integer-pair solution $(x,y)$. This one integer-pair solution though generates infinitely many other integer pair solutions via:
$$(x,y)\longrightarrow (x+mp,y+mn),\,m\in\mathbb{Z}\tag{2}$$
So there are infinitely many integer-pair solutions $(x,y)$ with $x$ and/or $y$ positive.
$\color{red}{\text{This paragraph is wrong. }}$ Now consider the integer-pair $(x,y)$ where $x$ is smallest, i.e $1\leq x < p$. Since $1\leq n < p$, we have that $1\leq xn < np$. But from eq. (1) this immediately implies $1\leq y < n <p$. $\blacksquare$
Corrected "Proof".
Starting from the $\color{red}{\text{"wrong" }}$ paragraph: Now consider the integer-pair $(x,y)$ where $y$ is within the range $1\leq y \leq n$. Multiplying this inequality by $p$, we see that:
$$p\leq yp \leq np \tag{3}$$
Since the integer solution $(x,y)$ solves the Diophantine equation (1), we can replace $yp$ in the above inequality with:
$$p \leq xn-1 \leq np \tag{4}$$
Adding $1$ to both sides, and then dividing by $n$ (which is allowed since $n$ is a positive integer):
$$\frac{p}{n}+\frac{1}{n} \leq x \leq p+\frac{1}{n}\tag{5}$$
At this point we realize the falsehood of the problem statement which @Servaes notified us about in his/her answer below. If $n=1$, we see that no integer $x$ can satisfy the inequality (5). In fact, only when $n>1$ does a solution exist, and even then the value of $x$ must lie within the range:
$$1< \lfloor \frac p n + \frac 1 n \rfloor \leq x \leq p $$
$$\implies \boxed{1 < x \leq p}$$
In contrast to the range proposed in the problem statement.