Problem Statement.

Let $n,p$ be positive integers where $p$ is a prime number and $n<p$. Consider the following linear Diophantine equation:

$$xn-yp=1tag{1}$$

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:

$$(x,y)longrightarrow (x+mp,y+mn),,minmathbb{Z}tag{2}$$

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 $1leq x < p$. Since $1leq n < p$, we have that $1leq xn < np$. But from eq. (1) this immediately implies $1leq y < n <p$. $blacksquare$

Corrected "Proof".

Starting from the $color{red}{text{"wrong" }}$ paragraph: Now consider the integer-pair $(x,y)$ where $y$ is within the range $1leq y leq n$. Multiplying this inequality by $p$, we see that:

$$pleq yp leq np tag{3}$$

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.

Your proof is a good start, but is lacking in some details: In the sentence

Now consider the integer-pair $(x,y)$ where $x$ is smallest, i.e $1leq x<p$.

You implicitly claim that such an $x$ always exists, but this requires an argument. This is not hard:

From your equation (2) it easily follows that there exists a pair $(x,y)$ with $0leq x<p$, for example by dividing $x$ by $p$ by means of the Euclidean algorithm. And indeed $xneq 0$ as otherwise
$$1=xn-yp=-yp,$$
which is impossible as $p$ is prime.$ square$

More seriously, in the sentence

But from eq. (1) this immediately implies $1leq y<n<p$.

This is false! From the inequalities $1leq xn<np$ and equation (1) you get
$$1leq1+yp<npqquadtext{ and so }qquad 0leq y<n-frac{1}{p}<n.$$
But you cannot exclude $y=0$ in general; you can only exclude this if $n>1$. If $n=1$ you have the solution $(x,y)=(1,0)$, and in fact a more careful look shows that the statement you are trying to prove is false when $n=1$.