Does there exist three non-zero homogeneous polynomials in $k[x,y]$, $k$ an algebraically closed field (ADDED: of characteristic 0), $f_1$ of degree 4, $f_2$ of degree 6, and $f_3$ of degree 3, and such that $f_1$ and $f_2$ do not have a common denominator of degree 2, that satisfy the equation $$ f_1^3 + f_2^2 = f_3^4\ ? $$
This elementary problem arises from the following question: does there exist a rational elliptic surface with a discriminant divisor that consists of three distinct points with multiplicity 4 --- which amounts to three singular fibres of the Kodaira $I_4$ type --- see Heckman-Looijenga, The moduli space of rational elliptic surfaces.
The following is a first draft of an answer; it might contain errors, I will double check and clean it up soon, perhaps tomorrow. It shows that no such polynomials exist if $\operatorname{char}k\neq2,3$.
If such homogeneous polynomials $f_1,f_2,f_3\in k[x,y]$ exist, then $$f_1^3=f_3^4-f_2^2=(f_3^2+f_2)(f_3^2-f_2).\tag{1}$$
The square of the $\gcd$ of the factors on the right hand side divides $f_1^3$, so $\gcd(f_3^2+f_2,f_3^2-f_2)=h^3$ for some monic $h\in k[x,y]$. Then $h^3\mid2f_2$ and $h^6\mid f_1^3$, so $h^2\mid f_1$. From your requirement that $f_1$ and $f_2$ do not have a quadratic common divisor it follows that $h=1$. It also follows that $\gcd(f_2,f_3)=1$.
It now follows from $(1)$ that both $f_3^2+f_2$ and $f_3^2-f_2$ are coprime cubes. Let $h_1,h_2\in k[x,y]$ be such that $$h_1^3=f_3^2+f_2 \qquad\text{ and }\qquad h_2^3=f_3^2-f_2.$$ Comparing degrees in identity $(1)$ and in the above shows that $\deg h_1=\deg h_2=2$. Then $$2f_2=(f_3^2+f_2)-(f_3^2-f_2) =h_1^3-h_2^3=(h_1-h_2)(h_1^2+h_1h_2+h_2^2),$$ $$2f_3^2=(f_3^2+f_2)+(f_3^2+f_2) =h_1^3+h_2^3=(h_1+h_2)(h_1^2-h_1h_2+h_2^2).\tag{2}$$
Of course $\gcd(h_1+h_2,h_1^2-h_1h_2+h_2^2)$ divides $$(h_1-2h_2)(h_1+h_2)-(h_1^2-h_1h_2+h_2^2)=-3h_2^2,$$ $$(h_2-2h_1)(h_1+h_2)-(h_1^2-h_1h_2+h_2^2)=-3h_1^2,$$ and because $\operatorname{char}{k}\neq3$ it follows that the $\gcd$ divides both $h_1^2$ and $h_2^2$, hence it also divides $$f_2=\tfrac12(h_1^3-h_2^3)\qquad\text{ and }\qquad f_3^2=\tfrac12(h_1^3+h_2^3),$$ and so it equals $1$. Then both factors of $2f_3^2$ at the right hand side of $(2)$ are squares, meaning that there exist coprime $u,v\in k[x,y]$ such that $$h_1+h_2=u^2\qquad\text{ and }\qquad h_1^2-h_1h_2+h_2^2=v^2.$$ Then $u$ and $v$ are again homogeneous and $u^2v^2=2f_3^2$, so $\deg u+\deg v=3$. Then $$\deg u^2\leq\max\{\deg h_1,\deg h_2\}=2 \qquad\text{ and }\qquad \deg v^2\leq\max\{2\deg h_1,2\deg h_2\}=4,$$ and from $\deg u+\deg v=3$ it follows that $\deg u=1$ and $\deg v=2$. A bit of algebra shows that \begin{eqnarray*} 3h_1h_2 &=&(h_1+h_2)^2-(h_1^2-h_1h_2+h_2^2)\\ &=&u^4-v^2\\ &=&(u^2+v)(u^2-v),\\ 3(h_1^2+h_1h_2+h_2^2) &=&3(h_1^2-h_1h_2+h_2^2)+6h_1h_2\\ &=&3v^2+2(u^2+v)(u^2-v)\\ &=&2u^4+v^2,\\ 3(h_1-h_2)^2 &=&3(h_1+h_2)^2-6h_1h_2\\ &=&3u^4-4(u^2+v)(u^2-v)\\ &=&-u^4+4v^2\\ &=&(2v+u^2)(2v-u^2). \end{eqnarray*} In the latter we see that $2v+u^2$ and $2v-u^2$ are coprime because $u$ and $v$ are, and so both are squares. Let $s,t\in k[x,y]$ be such that $$s^2=2v+u^2\qquad\text{ and }\qquad t^2=2v-u^2.$$ Note that both $s$ and $t$ are linear and homogeneous. Then $$4v=s^2+t^2\qquad\text{ and }\qquad 2u^2=s^2-t^2.$$ Now we can express $f_2^2$ and $f_3^4$ in terms of $s$ and $t$ as follows: \begin{eqnarray*} 2^{12}f_3^4 &=&2^{10}u^4v^4=(s^2-t^2)^2(s^2+t^2)^4\\ &=&t^{12}+2s^2t^{10}-s^4t^8-8s^6t^6-s^8t^4+2s^{10}t^2+s^{12},\\ 2^{10}3^3f_2^2 &=&2^{10}3^3(h_1-h_2)^2(h_1^2+h_1h_2+h_2^2)^2\\ &=&2^{10}(2v+u^2)(2v-u^2)(2u^4+v^2)^2\\ &=&(4v+2u^2)(4v-2u^2)(2^5u^4+2^4v^2)^2\\ &=&4s^2t^2(2^3(s^2-t^2)^2+(s^2+t^2)^2)^2\\ &=&4s^2t^2(9s^4-15s^2t^2+9t^4)^2\\ &=&36(9s^2t^{10}-30s^4t^8+43s^6t^6-30s^8t^4+9s^{10}t^2), \end{eqnarray*} and this allows us to also express $f_1^3$ in terms of $s$ and $t$ as follows: \begin{eqnarray*} 3(16f_1)^3 &=&3(2^{12}f_3^4)-2^4(2^83^1f_2^2)\\ &=&3t^{12}-138s^2t^{10}+477s^4t^8-712s^6t^6+477s^8t^4-138s^{10}t^2+3s^{12}. \end{eqnarray*} This is a palindromic polynomial in $s^2$ and $t^2$, which means that \begin{eqnarray*} (16f_1)^3&=& t^{12}-43s^2t^{10}+159s^4t^8-238s^6t^6+159s^8t^4-43s^{10}t^2+s^{12}\\ &=&(st)^6q\left(\left(\tfrac ts\right)^2+\left(\tfrac st\right)^2\right) \end{eqnarray*} for some cubic $q\in k[X]$. Then comparing degrees shows that $q$ must be the cube of a linear polynomial. A bit more algebra shows that \begin{eqnarray*} q(z+z^{-1}) &=&z^3-43z^2+159z-238+159z^{-1}-43z^{-2}+z^{-3}\\ &=&\left(z+z^{-1}\right)^3-43\left(z+z^{-1}\right)^2+156\left(z-z^{-1}\right)-152. \end{eqnarray*} But it is clear that $$q=X^3-43X^2+156X-152,$$ is not a cube in $k[X]$. We have finally reached a contradiction; no such polynomials exist.