Find all solutions in $\mathbb{N}$ to $a^a=a^b+b^a$

We already know that $a>b$. With this, we can prove that $$a^a>a^b+b^a,$$ thus proving that no solutions exist. We first prove a little lemma, and then we use induction on $a$.

Lemma: If $a>b$, $$a^{b+1}>(a+1)^b.$$

Proof: If $a=2$, then $b$ is necessarily equal to $1$, and the inequality holds, as $4>3$. Otherwise, we have $$a>e>\left(1+\frac1a\right)^a>\left(1+\frac1a\right)^b\Rightarrow$$ $$a^{b+1}>(a+1)^b,$$ as wanted. Here, $e\approx2.71828$ is Euler’s constant. $\square$

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With this, we begin our proof.

If $a=b+1$, the base case, we have $$(b+1)^b>b^b\Rightarrow$$ $$b\cdot(b+1)^b>b^{b+1}\Rightarrow$$ $$(b+1)^{b+1}>(b+1)^b+b^{b+1}.$$ Now, assuming $a^a>a^b+b^a$ for some $a>b$, we have $$(a+1)^{a+1}>a^{a+1}>a^{b+1}+a b^a>(a+1)^b+b^{a+1},$$ by our induction hypothesis and our lemma. This completes our proof. $\blacksquare$

As you noted, $a>b\geq 1$. Then

$$0=a^b+b^a-a^a\leq a^{a-1}+(a-1)^a-a^a$$

$$0\leq \left(1-\frac{1}{a}\right)^a+\frac{1}{a}-1=\left(1-\frac{1}{a}\right)^a-\left(1-\frac{1}{a}\right)$$

$$=\left(1-\frac{1}{a}\right)\left(\left(1-\frac{1}{a}\right)^{a-1}-1\right)<\left(1-\frac{1}{a}\right)\left(1-1\right)=0$$

As this is a contradiction, we conclude there are no positive integer solutions.

$a^a>b^a$ and so $a>b$. Let $a=b+d$.

Consider any prime $p$ dividing $a$ and let the maximum powers of the prime $p$ dividing $a$ and $b$ be $p^k$ and $p^l$, respectively. Then comparing the powers of $p$ dividing each side of $$a^b(a^d-1)=b^{b+d}$$ we obtain $bk=(b+d)l$.

Let $\frac{b}{b+d}=\frac{u}{v}$, where $u$ and $v$ are coprime. Then there is a positive integer $t$ such that $a=tv,b=tu$.

Also, there is a positive integer $s$ such that $k=sv,l=su$. Then $a$ is a $v$th power and so there is a positive integer $N$ such that $a=N^v$ and $b=N^uM$, where $N$ and $M$ are coprime.

Then the original equation cancels down to $$N^{vt(v-u)}-1=M^{tv}. $$ By FLT we have $tv\le2$ i.e $a\le2$. There are no solutions.