Find all solutions in positive integers to $a^b -b^a = 3$ 
Find all solutions in positive integers to $a^b -b^a = 3$

It appears that the only solution is $a=4, b=1$. 
Modular arithmetic is not of much help (apart from deriving that $a,b$ must have opposite parity). 
We know that $a^x$ dominates the polynomial $x^a$ but this does not seem to yield the result. 
Tried hard to estimate a bound for $a^b - b^a$ but with no success. 
Any help is appreciated. This is a problem from a regional mathematics Olympiad.
Added on 26 March, 2017:
I was trying to find a solution that involves no Calculus (or a minimum of Calculus) since this is a problem from a Junior Olympiad in which the students do not have a knowledge of Calculus. The following argument seems to work. Please point out flaws/mistakes, if any.
Lemma 1 For any $n \geq 4$, $$n^{n+1} - (n+1)^n > (n-1)^n - n^{n-1}$$
 Proof 
\begin{align*}
&\qquad n^{n+1} - (n+1)^n > (n-1)^n - n^{n-1} \\
&\Leftrightarrow n^{n+1} + n^{n-1} > (n-1)^n + (n+1)^n \\
&\Leftrightarrow n+\frac{1}{n} > \left(1-\frac{1}{n}\right)^n + \left(1+\frac{1}{n}\right)^n
\end{align*}
Since $\left(1+\frac{1}{n}\right)^n < 3$ and $\left(1-\frac{1}{n}\right)^n < 1$, we have 
$$\left(1+\frac{1}{n}\right)^n+\left(1-\frac{1}{n}\right)^n < 4 < n +\frac{1}{n}$$
 Lemma 2 For any $k$ and $n \geq 4$, 
$$n^{n+k} -(n+k)^n  > n^{n+1} - (n+1)^n$$
Proof 
\begin{align*}
&\qquad n^{n+k} -(n+k)^n  > n^{n+1} - (n+1)^n \\
&\Leftrightarrow n^{n+k} -n^{n+1} > (n+k)^n - (n+1)^n \\
&\Leftrightarrow n^k - n > \left(1+\frac{k}{n}\right)^n - \left(1+\frac{1}{n}\right)^n 
\end{align*}
Since $\left(1+\frac{k}{n}\right)^n < 3^k$ and $\left(1-\frac{1}{n}\right)^n < 1$, we have 
$$\left(1+\frac{k}{n}\right)^n+\left(1-\frac{1}{n}\right)^n < 3^k +1  < n^k - n $$
as $n > 4$.
Now suppose that $a^b - b^a = 3$.  Let $a \geq 4$. If $b = a+k > a$, then we have 
\begin{align*}
a^{a+k} - (a+k)^a &> a^{a+1} - (a+1)^a \\
&> (a-1)^a - a^{a-1} \\
&> \cdots \\
&> 3^4 - 4^3 = 17
\end{align*}
Thus there are no solutions with $b > a \geq 4$. 
If $a > b \geq 4$, then 
$$a^b - b^a = -(b^a - a^b) \leq -17 $$
from what we have seen above. Thus there are no solutions if $a > b \geq 4$. Thus all solutions can be only in the range $1 \leq a,b \leq 4$. Clearly, in this range only $a=4, b=1$ satisfies the given equation.

Consider $$f_a(x)=x\ln(a)-a\ln(x)$$ with $a,x\ge 3$
We have $$f_a'(x)=\ln(a)-\frac{a}{x}>0$$ for $x>\frac{a}{\ln(a)}$
In particular, $f_a(x)$ is strictly increasing for $x\ge a$ and we have $$f_a(a+1)=(a+1)\ln(a)-a\ln(a+1)$$
The function $g(x)=(x+1)\ln(x)-x\ln(x+1)$ has derivate $g'(x)=\ln(x)-\ln(x+1)+\frac{x+1}{x}-\frac{x}{x+1}=\ln(\frac{x}{x+1})+\frac{2x+1}{x^2+x}>\ln(\frac{x}{x+1})+\frac{1}{x}=\frac{1}{x}-\ln(1+\frac{1}{x})>0$ for $x\ge 3$
So, we have $f_a(x)\ge 0.235566$ for $3\le a\le x$ hence $$x\ln(a)\ge a\ln(x)+0.235566$$ implying $$a^x>1.26x^a$$
Since $x^a\ge 27$, we can conclude $$a^x>x^a+7$$
This rules out $3\le a<x$
The case $3\le x<a$ can be ruled out analogue by changing the roles of $a$ and $x$. $a=x$ can be rules out anyway.
Now, we show $2^a>a^2+3$ for $a\ge 5$ by induction
$a=5$ : $2^5=32>28=5^2+3$
$$2^{a+1}=2\cdot 2^a>2(a^2+3)=2a^2+6>a^2+2a+4=(a+1)^2+3$$
for $a^2>2a-2$ , or $a^2-2a+2=(a-1)^2+1>0$ , which is true for every $a$
The case $x=1$ and the remaining cases for $x=2$ can be verified easily.
