Finding $f(x)$ given a function $f : \Bbb N \to \Bbb N$ which satisfies $f\circ f(x) + f(x) = 2x+15$

A reasonable thing to guess is that $f(x)$ is a polynomial. However, if the leading term is $x^n$, you get a leading term on the LHS of $x^{n^2}$ vs the leading term on the RHS of $x$. This means $n=1$ if a polynomial is going to work. So then we try $f(x)=ax+b$ and solve for the unknown coefficients.

We have the simple bound $$f(x)\le f(f(x))+f(x)= 2x+15.$$ Also, $$\begin{array}{}f(f(f(x)))+2x+15&=f(f(f(x)))+f(f(x))&+f(x)\hphantom{,}\\&=2f(x)+15&+f(x),\end{array}$$ which implies $$\tag13f(x)-2x=f(f(f(x))).$$ Let $$ C=\{\,c\in\Bbb R\mid \forall x\in\Bbb N\colon f(x)\ge cx\,\}$$ Clearly $0\in C$ and $C$ is bounded from above. By $(3)$. for any $0\le c\in C$, we have $f(x)\ge\frac{2x+f(f(f(x)))}{3}\ge \frac{2+c^3}{3}x$ for all $x\in\Bbb N$, i.e., also $\frac{2+c^3}{3}\in C$. Let $\gamma=\sup C$. Then $\gamma\ge\frac{2+\gamma^3}{3}$, which (using the factorization $\frac{X^3+2}3-3X=\frac13(X-1)^2(X+2)$ and that $\gamma\ge 0$) implies $\gamma\ge1$. Thus $f(x)>cx$ for all $c<1$, i.e., $$\tag2f(x)\ge x.$$ Thus the function $g(x)= f(x)-x$ is bounded from below. Let $$A=\{\,a\in\Bbb R\mid \forall x\in\Bbb N\colon f(x)\ge x+a\,\}$$ and $$B=\{\,b\in\Bbb R\mid \forall x\in\Bbb N\colon f(x)\le x+b\,\}.$$ As seen, $0\in A$. Note that $$g(f(x))+2g(x) =f(f(x))+f(x)-2x=15$$ so $a\in A$ implies that $g(x)\le\lfloor \frac{15-a}2\rfloor$ for all $x$, i.e., $\lfloor \frac{15-a}2\rfloor\in B$. Similarly, $b\in B$ implies $g(x)\ge\lceil\frac{15-b}{2}\rceil$, i.e., $\lceil\frac{15-b}{2}\rceil\in A$. Thus we conclude $$0\in A\implies 7\in B\implies 4\in A\implies 5\in B\implies 5\in A. $$ We conclude that $g(x)=5$ for all $x$, and thus $$f(x)=x+5. $$

The given answers (all brilliant) are sufficient for answering the OPs question, however I would like to point out another approach which often works for determining all solutions of a functional equation similar to the one in question:

Let $a\in\mathbb{N}$ and define the sequence $\{a_n\}_{n\ge0}\subset\mathbb{N}$ by $a_0=a$ and $a_{n+1}=f(a_n)$. Using the condition we obtain the recurrence relation $$ a_{n+2}+a_{n+1}=2a_n+15. $$ In order to apply the usual procedure of solving such a linear recurrence, we have to get rid of the $+15$-term. This can be done by introducing $\{b_n\}_{n≥0}$ as $a_n=b_n+5n$; the $b_n$ then satisfy $$ b_{n+2}+b_{n+1}=2b_n. $$ The characteristic polynomial of this relation is $p(\lambda)=\lambda^2+\lambda-2=(\lambda-1)(\lambda+2)$. Thus we have $$ b_n=x+y(-2)^n\qquad\forall n\in\mathbb{N}_0 $$ for some constants $x,y\in\mathbb{R}$. But now if $y\neq 0$, there exists a $n\in\mathbb{N}$ such that $a_n<0$ (because we will have $|y|2^N>x+5N$ for large enough $N$), which contradicts the fact that $f:\mathbb{N}\to\mathbb{N}$. Thus $y=0$ and therefore $\{b_n\}$ is a constant sequence. As $b_0=a_0=a$ we conclude $$ b_n=a\qquad\forall n\in\mathbb{N}_0 $$ and hence $$ a_n=a+5n\qquad\forall n\in\mathbb{N}_0\\ \implies f(a)=a_1=a+5. $$ As $a\in\mathbb{N}$ was arbitrary, we conclude $$ f(n)=n+5\qquad\forall n\in\mathbb{N}. $$