We assume that $ x\ne a$. Let $ a=\prod_{i=1}^{k} {p_i^{x_i}}$, $ b=\prod_{i=1}^k {p_i^{y_i}}$, and $ x=\prod_{i=1}^{k} {p_i^{z_i}}$. Henceforth, we will ignore simply denote $ \prod_{i=1}^k$ as $ \prod$ for simplicity. Furthermore, when we refer to $ y_i$, we do not mean $ y_i$ collectively, but for rather for a fixed $ y_i$ (or else otherwise noted.) Then, \[ x^{a+b}=a^bb\iff \prod {p_i^{z_i\cdot (\prod {p_i^{y_i}}+\prod {p_i^{x_i}})}=\prod {p_i^{y_i+x_i\cdot \prod {p_i^{y_i}}}}\iff}\] \[ x_i\cdot \prod {p_i^{y_i}}+y_i=z_i\cdot (\prod {p_i^{x_i}}+\prod {p_i^{y_i}})\] for all $ 1\le i\le k$. Now, if $ y_i<x_i$, then $ p_i^{y_i}|\prod {p_i^{x_i}}$, and $ p_i^{y_i}$ is a factor of all the other terms in the equation except $ y_i$. This means that $ p_i^{y_i}|y_i$, which is only possible if $ y_i=0$ since $ y_i<p_i^{y_i}$. If $ y_i=0$, then we can obtain that $ (x_i-z_i)\cdot \prod {p_i^{y_i}}=\prod {p_i^{x_i}}$. Yet, $ p_i$ is not a factor of $ \prod {p_i^{y_i}}$ when it is a factor of $ \prod {p_i^{x_i}}$. This means that $ p_i^{x_i}$ is a factor of $ x_i-z_i<x_i<p_i^{x_i}$, so $ x_i-z_i=0$, so $ \prod {p_i^{x_i}}=0$, which is a contradiction. This means that $ x_i\le y_i$ for all $ 1\le i\le k$, so $ \prod {p_i^{x_i}}|\prod {p_i^{y_i}}$, meaning that $ \prod {p_i^{x_i}}|y_i$, so $ a|y_i$ for all $ 1\le i\le k$. Letting $ y_i=ak_i$, we have that $ ak_i+x_ib=z_ia+z_ib$, so $ a(k_i-z_i)=b(z_i-x_i)$, so \[ \frac{k_i-z_i}{z_i-x_i}=\frac{b}{a}=\prod {p_i^{ak_i-x_i}}\] Now, since $ ak_i=y_i$, which is a multiple of $ a>x_i$, we have that $ ak_i>x_i$, meaning that $ \prod {p_i^{ak_i-x_i}}>ak_i-x_i\ge k_i-x_i$. If $ k_i-z_i<0$, then if $ z_i-x_i>0$, we would have that the left hand side of the equation is negative when the other side is positive, giving us a contradiction. Then, $ z_i-x_i<0$, so $ \frac{k_i-z_i}{z_i-x_i}=\frac{z_i-k_i}{x_i-z_i}<z_i-k_i+x_i-z_i<x_i$. Yet, $ a\ge p_i^{x_i}\ge 2x_i$, meaning that \[ \prod {p_i^{ak_i-x_i}}>ak_i-x_i\ge a-x_i\ge x_i>\frac{k_i-z_i}{z_i-x_i}\] which is a contradiction. Otherwise, we have that $ k_i-z_i, z_i-x_i>0$, so \[ \frac{k_i-z_i}{z_i-x_i}<k_i-z_i+z_i-x_i<\prod {p_i^{ak_i-x_i}}\] which is a contradiction, so we conclude that $ z_i=x_i$ for all $ 1\le i\le k$, so $ a=x$. This is a contradiction, so we get that $ a=x$. This means that $ x^{x+b}=x^b\cdot b$, so $ b=x^x$, as desired.

What??? $ x^b|a^bb$ thus $ x|a$ or some $ b$-th power divides $ b$. We toss out second case as $ n^b\geq 2^b>b$ Thus $ x|a$. Suppose $ a\geq 2x$; then $ b=x^a(x/a)^b\geq x^a 2^b>b$. Thus $ a=x$ and the result follows.

Step 1 $x\mid a$. For each prime $p$ we have $bv_p(x)<(a+b)v_p(x)=bv_p(a)+v_p(b)<b(v_p(a)+1)$, so $v_p(x)<v_p(a)+1$, hence $x\mid a$. Step 2 $a\mid b$. Write $a=cx$, then $x^{cx+b}=(cx)^bb$ reduces to $x^a=c^bb$. For each prime $p\mid a$ we have $av_p(x)=bv_p(c)+v_p(b)$. Let $v_p(a)=k\geq 1$, then $v_p(av_p(x))\geq k$, therefore $v_p(bv_p(c)+v_p(b))\geq k$. If $v_p(b)=0$ - If $v_p(c)=0$, then $av_p(x)=0$, so $v_p(x)=0$, but this is impossible as $p\mid a\to p\mid x$ - Else, $v_p(v_p(c))=v_p(bv_p(c)+v_p(b))\geq k=v_p(a)\geq v_p(c)$, which is impossible. If $0<v_p(b)<k$, then $v_p(v_p(b))\geq \min(v_p(bv_p(c)+v_p(b)),v_p(bv_p(c)))$. However, since $v_p(v_p(b))<v_p(b)<k\leq v_p(bv_p(c)+v_p(b)$ and $v_p(v_p(b))<v_p(b)\leq v_p(bv_p(c))$, this is also impossible. Therefore we have $v_p(b)\geq v_p(a)$ for all prime $p$, thus $a\mid b$. Step 3 $a=x,b=x^x$. For each prime $p$, from $a\mid b$ and $(a+b)v_p(x)=bv_p(a)+v_p(b)$, it follows that $a\mid v_p(b)$ for all prime $p$. Therefore $b=d^a$ for some $d\in\mathbb{N}$. If $b=1$ it's obvious that $a=x=1$, so suppose that $b\geq 2$, so that $d\geq 2$. Thus for each prime $p$, $(a+d^a)v_p(x)=d^av_p(a)+av_p(d)$. This gives $v_p(x)=v_p(a)+\frac{av_p(d)-av_p(a)}{a+d^a}$. Now, note that $av_p(a)-av_p(d)\leq av_p(a)<a^2<a+2^a<a+d^a$ for all $a$, so $v_p(x)>v_p(a)-1$. However, from $x\mid a$, $v_p(x)\leq v_p(a)$, so $v_p(x)=v_p(a)$ for all prime $p$. Therefore $a=x$, and thus $b=x^x$.